How to prove exponential correspondence for tensors?

Question

Let ${\rm Vec}_{\mathbb{R}}$ denote the category of finite dimensional vector spaces over the real numbers. Let ${\rm Hom}_k(V,W)$ denote the space of multilinear maps $\underbrace{V\times\dots\times V}_{k\ times} \to W$ and $V_k^1$ denote the space of multilinear maps $\underbrace{V\times\dots\times V}_{k\ times} \times V^* \to \mathbb{R}$.

I want to show that there is a natural isomorphism between $Hom_k(V,V)$ and $V_k^1$. I have reduced the problem to showing the existence of a natural isomorphism between $Hom_k(V,V^{**})$ and $V_k^1$. Intuitively it is clear to me that such an identification can be accomplished by currying, that this is a vector space isomorphism should be verifiable simply by checking the definitions, and that naturality should follow from (some variant of) the Yoneda Lemma. However, I am having some difficulty finding the endofunctors in $Vec_{\mathbb{R}}$ that would allow me to verify all of these claims.

Reduction: I know that, given a finite dimensional vector space $V$, there is a natural isomorphism with $V^{**}$. Using hom functors, it isn't that difficult to extend this to a natural isomorphism between $Hom_k(V,V)$ and $Hom_k(V,V^{**})$.

What I have tried: I have tried to find functors $$f_1: Vec_{\mathbb{R}} \to Vec_{\mathbb{R}}, V \mapsto Hom_k(V,V^{**})$$ and $$f_2: Vec_{\mathbb{R}} \to Vec_{\mathbb{R}}, V\mapsto V_k^1.$$ However, given vector space homomorphisms $T:V\to W$, it is unclear to me how to find vector space homomorphisms $Hom_k(V,V^{**}) \to Hom_k(W,W^{**})$ or $Hom_k(W,W^{**}) \to Hom_k(V,V^{**})$ which would give the morphism part of the functors. First I tried using a contravariant hom functor, i.e. pre-composition by T, $$T \mapsto f(T(\cdot),\dots,T(\cdot)),$$ but the problem with this is that it only maps $Hom_k(W,W^{**}) \to Hom_k(V,W^{**})$, but clearly I need a map $Hom_k(W,W^{**}) \to Hom_k(V,V^{**})$.

So then I tried restricting from $Vec_{\mathbb{R}}$ to the subcategory $\mathbb{V}$, defined to be the category whose objects are all $n$-dimensional vector spaces over $\mathbb{R}$ and whose morphisms consist only of invertible linear transformations, and then finding functors $$f_1: \mathbb{V} \to Vec_{\mathbb{R}}, \quad f_2: \mathbb{V} \to Vec_{\mathbb{R}},$$ with the same rules of assignment for the objects as before, namely $f_1: V \mapsto Hom(V,V^{**})$ and $f_2: V \mapsto V_k^1$. Then I tried the following rule of assignment for morphisms for $f_1$: $$T\mapsto T^{-1} f(T(\cdot),\dots,T(\cdot))$$ -- this does map $Hom_k(W,W^{**})\to Hom_k(V,V^{**})$ as desired, but it turns out this satisfies neither associativity nor functoriality. Now I am stuck. I haven't tried to find a rule of assignment for morphisms for $f_2$, but it seems like one would run into the analogous problems.

However, without two endofunctors $f_1$ and $f_2$, I can't show that there is a natural transformation between them which is also an isomorphism and which is a vector space isomorphism when applied to objects.

In terms of exponential objects, I think what I want to show is equivalent to: $$ V^{V^k} \cong \mathbb{R}^{V^k \otimes V^*} .$$ Then what I have reduced it to is showing $$ V^{V^k} \cong (\mathbb{R}^{V^*})^{V^k} \cong \mathbb{R}^{V^k \otimes V^*} .$$ Now the left isomorphism is essentially easy, since $V \cong V^{**}:=\mathbb{R}^{V^*}$, so what I need to show is the right isomorphism, which is the vector space version of the natural exponential object correspondence. So if I can prove that $(X^Y)^Z \cong X^{Y \times Z}$ in $Set$, then maybe the proof will carry over almost word for word for the corresponding version in $Vec_{\mathbb{R}}$. I still need to look into this line of approach more, although so far I have not succeeded yet in demonstrating the naturality of the isomorphism $(X^Y)^Z \cong X^{Y \times Z}$ in $Set$, so it might not be as easy as I am hoping.

EDIT: If I can show that $Vec_{\mathbb{R}}$ is Cartesian closed, then this should follow from Exercise 1(b), section 6.6., p.123, of Awodey's Category Theory. Although the exercise is only to prove that we have an isomorphism, $(A^B)^C \cong A^{B \times C}$, but does not include a proof of naturality.

Answer 1

Your "functors" aren't actually functors. I'll stick to standard notations because "$\operatorname{Hom}_k(V,V)$" is too easy to mistake for the space of $k$-linears maps $V \to V$ for some field $k$... I'll also let the base field $\mathbb{R}$ be implied throughout.

Let's do the simplest case, $k=1$. Then you have a bifunctor $$\operatorname{Hom} : \mathsf{Vec}^{\mathrm{op}} \times \mathsf{Vec} \to \mathsf{Vec}, \; (V,W) \mapsto \operatorname{Hom}(V,W)$$ It's not possible to compose that with the functor $\mathsf{Vec} \to \mathsf{Vec} \times \mathsf{Vec}$, simply because the (co)domains don't match... So the mapping $V \mapsto \operatorname{Hom}(V,V)$ doesn't yield a functor a priori. I can't see any reasonable way to make that into a functor. Your question is doomed from the start...

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However you do have the bifunctor $\operatorname{Hom}$ as above. You also have another bifunctor, say $$\Phi : \mathsf{Vec}^{\mathrm{op}} \times \mathsf{Vec} \to \mathsf{Vec}, \; (V,W) \mapsto \operatorname{Hom}(V \otimes W^*, \mathbb{R}).$$

This is indeed a bifunctor; given $f : V' \to V$ and $g : W \to W'$, you get $\Phi(f,g) : \Phi(V,W) \to \Phi(V',W')$ given by $t \mapsto t \circ (f \otimes g^*)$ (where $g^* : \operatorname{Hom}(W', \mathbb{R}) \to \operatorname{Hom}(W,\mathbb{R})$ is precomposition by $g$).

These two functors are naturally isomorphic, when you restrict to finite-dimensional spaces. Indeed define a natural transformation $\eta : \operatorname{Hom} \to \Phi$ by: \begin{align} \eta_{(V,W)} : \operatorname{Hom}(V,W) & \to \operatorname{Hom}(V \otimes W^*, \mathbb{R}) \\ t & \mapsto (\eta(t) : v \otimes \psi \mapsto \psi(t(v))) \end{align}

It's not hard (but it's a bit tedious) to check that this is a natural transformation. Both spaces have the same dimension, namely $(\dim V) (\dim W)$. So to check that this is an isomorphism for all $(V,W) \in \mathsf{Vec}^\mathrm{op} \times \mathsf{Vec}$, it suffices to check that this is surjective.

For a fixed couple $(V,W)$, let $(w_1, \dots, w_n)$ be a basis of $W$ (recall that it's finite dimensional), and let $(w_1^*, \dots, w_n^*)$ be the dual basis of $W^*$. Suppose given some $\beta \in \operatorname{Hom}(V \otimes W^*, \mathbb{R})$. Then you can let $\alpha \in \operatorname{Hom}(V, W)$ be defined by $$\alpha(v) = \sum_{i=1}^n \beta(v \otimes w_i^*) \cdot w_i.$$

It is now a very tedious (but completely mechanical) check that $\eta_{(V,W)}(\alpha) = \beta$. The mapping is surjective between spaces of the same finite dimension, hence it's an isomorphism.

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Finally, to get the general case, consider the functor $\bigotimes : \mathsf{Vec}^{\times k} \to \mathsf{Vec}$ given by $\bigotimes(V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k$, and compose it with the natural isomorphism $\eta$ to get a natural isomorphism $$\operatorname{Hom}(V_1 \otimes \dots \otimes V_k, W) \xrightarrow{\cong} \operatorname{Hom}(V_1 \otimes \dots \otimes V_k \otimes W^*, \mathbb{R}).$$

But again this is contravariant in the $V_i$ and covariant in $W$, so you can compose with $V \mapsto (V, \dots, V)$ because this is covariant in everything...

Answer 2

$\newcommand{\Vec}{\mathsf{Vec}}$$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\id}{\operatorname{id}}$This is for my future reference so I never have to write out these proofs again.

Claim: $\Hom(-,-):\Vec^{op}\times \Vec \to \Vec$ is a functor.

(a) Given $V,W \in \Vec$, we define the object part of $\Hom$ as follows: $$(V,W)\mapsto \Hom(V,W) $$ where $\Hom(V,W)$ is defined to be the space of all linear transformations $S: V \to W$.

(b) Given $(V,W) \overset{(f,g)}{\to} (V',W')$, i.e. $f: V' \to V$, $g: W \to W'$ linear transformations, we define the morphism part of $\Hom$ as follows: $$\Hom(f,g):S \mapsto g \circ S \circ f .$$ Note that $g \circ S \circ f \in \Hom(V',W')$, hence $\Hom(f,g): \Hom(V,W) \to \Hom(V'W')$ as required.

(c) We now show that $\Hom$ preserves identity morphisms. Let $(\id_V, \id_W)= \id_{(V,W)}: (V,W) \to (V,W)$ be the identity morphism for $(V,W)$. Then we want to show that $\Hom(\id_V,\id_W)$ is the identity morphism for $\Hom(V,W)$. So again let $S \in \Hom(V,W)$, then by definition: $$\Hom(\id_V, \id_W): S \mapsto \id_W \circ S \circ \id_V =S .$$ Thus $\Hom(\id_V,\id_W)=\id_{\Hom(V,W)}$, which is what we wanted to show.

(d) Given $(V,W) \overset{(f_1,g_1)}{\to} (V',W')$ and $(V',W') \overset{(f_2,g_2)}{\to} (V'',W'')$, i.e. linear transformations $f_2: V'' \to V', f_1: V' \to V, g_1: W \to W', g_2: W' \to W''$. First we compute $$\Hom((f_2,g_2)\circ(f_1, g_1) )=\Hom( f_1 \circ f_2, g_2 \circ g_1): S \mapsto (g_2 \circ g_1) \circ S \circ (f_1 \circ f_2). $$ Note that $(g_2 \circ g_1) \circ S \circ (f_1 \circ f_2): V'' \to W''$ is a linear transformation, i.e. $(g_2 \circ g_1) \circ S \circ (f_1 \circ f_2) \in \Hom(V'',W'')$ as expected and required. Secondly we compute: $$\Hom(f_2,g_2)\circ\Hom(f_1,g_1):S \mapsto g_2 \circ (g_1 \circ S \circ f_1) \circ f_2 .$$ Again, $g_2 \circ (g_1 \circ S \circ f_1) \circ f_2: V'' \to W''$ is a linear transformation, i.e. an element of $\Hom(V'',W'')$. And by associativity of function composition we clearly have $$(g_2 \circ g_1) \circ S \circ (f_1 \circ f_2) = g_2 \circ (g_1 \circ S \circ f_1) \circ f_2, $$ from which it follows that $$\Hom((f_2,g_2)\circ(f_1,g_1))=\Hom(f_2,g_2)\circ\Hom(f_1,g_1), $$ i.e. $\Hom$ does commute with composition of morphisms as required. Thus $\Hom$ is a functor.

Claim: $\Hom$ is a bifunctor.

This follows from the fact that $\Hom$ is a functor whose domain is a product category, $\Vec^{op}\times \Vec$.

Claim: $\Phi:\Vec^{op}\times\Vec \to \Vec$ is a functor.

(a) Given $V,W \in \Vec$, we define the object part of $\Phi$ to be: $$\Phi:(V,W)\mapsto\Hom(V\otimes W^*, \mathbb{R}).$$

(b) Given $(V,W) \overset{(f,g)}{\to} (V',W')$, i.e. $f: V' \to V, g: W \to W'$ linear transformations, we define the morphism part of $\Phi$ to be: $$\Phi(f,g): \sigma \mapsto \sigma \circ (f \otimes g^*), $$ with $\sigma \in \Hom(V \otimes W^*, \mathbb{R})$, $g^*: (W')^*=\Hom(W',\mathbb{R}) \to \Hom(W, \mathbb{R})=W^*$, $g^*: \tau \mapsto \tau \circ g\ \forall \tau\in(W')^*$, $(f \otimes g^*): V' \otimes (W')^* \to V \otimes W^*$, since $f: V' \to V$, $g: W \to W' \implies g^* : (W')^* \to W^*$ (adjoint/pre-composition is contravariant), thus $$V' \otimes (W')^* \overset{(f \otimes g^*)}{\to } V \otimes W^* \overset{\sigma}{\to} \mathbb{R}, $$ thus $\sigma \circ (f \otimes g^*): V' \otimes (W')^* \to \mathbb{R}$ as required, and thus $\Phi(f,g): \Phi(V,W)= \Hom(V \otimes W^*, \mathbb{R}) \to \Hom(V' \otimes (W')^*, \mathbb{R}) = \Phi(V',W')$ as claimed.

(c) We now show that $\Phi$ preserves identity morphisms. Let $(\id_V, \id_W)=\id_{(V,W)}$ be the identity morphism for $(V,W) \in \Vec^{op} \times \Vec$. Then we want to show that $\Phi(\id_v, \id_W)=\id_{\Phi(V,W)}=\id_{\Hom(V \otimes W^*, \mathbb{R})}$. Given $\sigma \in \Hom(V \otimes W^*, \mathbb{R})$, we have that $$\Phi(\id_V, \id_W): \sigma \mapsto \sigma \circ (\id_V \otimes \id_W^*), $$ now since $(\id_V,\id_W):(V,W)\to(V,W)$, we have that $(\id_V \otimes \id_W^*):V \otimes W^* \to V \otimes W^*$, so that $\sigma \circ (\id_V \otimes \id_W^*) \in \Hom(V \otimes W^*, \mathbb{R})$. We have by definition of tensor product of linear maps that for all $u \otimes \tilde{w} \in V \otimes W^*$: $$(\id_V \otimes \id_W^*)(u\otimes\tilde{w})=(\id_V(u))\otimes(\id_W^*(\tilde{w}))=u \otimes (\tilde{w} \circ \id_W) = u \otimes \tilde{w}, $$ in other words $(\id_V \otimes \id_W^*)=\id_{V \otimes W^*}$, so $\sigma \circ (\id_V \otimes \id_W^*) = \sigma \circ \id_{V \otimes W^*}= \sigma$, so in conclusion $\forall \sigma \in \Hom(V \otimes W^*, \mathbb{R}), \quad \Phi(\id_V,\id_W): \sigma \mapsto \sigma$, thus $\Phi(id_V,\id_W)=\id_{\Hom(V \otimes W^*, \mathbb{R})} = \id_{\Phi(V,W)}$.

(d) We now show that $\Phi$ commutes with composition. Given $(V,W) \overset{(f_1,g_1)}{\to}(V',W') \overset{(f_2,g_2)}{\to}(V'',W'')$, i.e. linear transformations $f_2: V'' \to V'$, $f_1: V' \to V$, $g_1: W \to W'$, $g_2: W' \to W''$. First we compute $$ \Phi( (f_2,g_2)\circ(f_1,g_1) ) = \Phi( f_1 \circ f_2, g_2 \circ g_1 ): \sigma \mapsto \sigma \circ ( (f_1 \circ f_2) \otimes (g_2 \circ g_1)^* ) = \sigma \circ ( (f_1 \circ f_2) \otimes (g_1^* \circ g_2^*) ) $$ Then we compute $$ \Phi(f_2,g_2)\circ\Phi(f_1,g_1):\sigma \mapsto (\sigma \circ (f_1 \otimes g_1^*))\circ (f_2 \otimes g_2^*) = \sigma \circ (f_1 \otimes g_1^*) \circ (f_2 \otimes g_2^*). $$ Now $f_1 \circ f_2: V'' \to V$, $(g_2 \circ g_1): W \to W'' \implies (g_2 \circ g_1)^*: \Hom(W'', \mathbb{R}) \to \Hom(W, \mathbb{R})$, $(g_2 \circ g_1)^*: \tau \mapsto \tau \circ (g_2 \circ g_1) = (\tau \circ g_2)\circ g_1 = (g_1^* \circ g_2^*) (\tau)$. Furthermore $$ (f_2 \otimes g_2^*): V'' \otimes (W'')^* \to V' \otimes (W')* $$ and $$ (f_1 \otimes g_1^*): V' \otimes (W')^* \to V \otimes W^* $$ thus $$ (f_1 \otimes g_1^*) \circ (f_2 \otimes g_2^*): V'' \otimes (W'') \overset{(f_2 \otimes g_2^*)}{\to} V' \otimes (W')^* \overset{(f_1 \otimes g_1^*)}{\to} V \otimes W^* $$ in other words $$ (f_1 \otimes g_1^*) \circ (f_2 \otimes g_2^* ): V'' \otimes (W'')^* \to V \otimes W^*. $$ Similarly, since $(f_1 \circ f_2): V'' \to V$ and $g_1^* \circ g_2^*: (W'')^* \to W^*$, we have by the definition of tensor product of maps that $$ (f_1 \circ f_2) \otimes (g_1^* \circ g_2^* ): V'' \otimes (W'')^* \to V \otimes W^*, $$ hence it is at least plausible that the two maps are equal.

Let $(u, \tilde{w}) \in V'' \otimes (W'')^*$. Then $$ (f_1 \otimes g_1^*) \circ (f_2 \otimes g_2^*) (u, \tilde{w}) = (f_1 \otimes g_1^*)( f_2(u) \otimes g_2^*(\tilde{w}) ). $$ Now since $f_2: V'' \to V'$, we have that $f_2(u) \in V'$ and since $g_2^*: (W'')^* \to (W')^*$, we have that $g_2^*(\tilde{w}) \in (W')^*$, thus $f_2(u) \otimes g_2^*(\tilde{w}) \in V' \otimes (W')^*$, which is the domain of $(f_1 \otimes g_1^*)$, so we can apply it, using once again the definition of tensor product of maps: $$ (f_1 \otimes g_1^*)( f_2(u) \otimes g_2^*(\tilde{w}) ) = f_1(f_2(u)) \otimes g_1^*(g_2^*(\tilde{w})) = (f_1 \circ f_2)(u) \otimes (g_1^* \circ g_2^*)(\tilde{w}) = ( (f_1 \circ f_2) \otimes (g_1^* \circ g_2^*) )(u, \tilde{w}). $$ In other words, we have shown that, for all $(u,\tilde{w}) \in V'' \otimes (W'')^*$, we have $$ [(f_1 \otimes g_1^*) \circ (f_2 \otimes g_2^*)](u, \tilde{w}) = [(f_1 \circ f_2) \otimes (g_1^* \circ g_2^*)](u, \tilde{w}), $$ i.e. that $(f_1 \otimes g_1^*) \circ (f_2 \otimes g_2^*) = (f_1 \circ f_2) \otimes (g_1^* \circ g_2^*)$ (i.e. that $\circ$ and $\otimes$ play nicely with each other). Anyway this has as an immediate consequence that $$ \Phi((f_2,g_2)\circ(f_1,g_1))(\sigma) = \Phi(f_1 \circ f_2, g_2 \circ g_1)(\sigma) = \\ \sigma \circ [(f_1 \circ f_2) \otimes (g_1^* \circ g_2^*) ] = \sigma \circ [ (f_1 \otimes g_1^*) \circ (f_2 \otimes g_2^*) ] = \Phi(f_2,g_2) \circ \Phi(f_1, g_1) $$ for all $\sigma \in \Hom(V \otimes W^*, \mathbb{R})$, i.e. that $\Phi( (f_2,g_2)\circ(f_1,g_1) ) = \Phi(f_2,g_2)\circ \Phi(f_1,g_1)$. Thus $\Phi$ commutes with composition, i.e. is functorial, and thus is a functor, as claimed.

Claim: $\Phi$ is a bifunctor.

This follows from the fact that $\Phi$ is a functor whose domain is a product category, $\Vec^{op}\times \Vec$.

Claim: There exists a natural transformation $\eta$ between $\Hom$ and $\Phi$.

A natural transformation $\eta$ is a family of morphisms in the target category ($\Vec$) indexed by the elements of the source category ($\Vec^{op} \times \Vec$), $\eta_{(V,W)}: \Hom(V,W) \to \Phi(V,W)$, such that, given $(V,W) \overset{(f,g)}{\to} (V',W')$, i.e. linear transformations $f: V' \to V$ and $g: W \to W'$, the following relationship holds: $$ \eta_{(V',W')} \circ \Hom(f,g) = \Phi(f,g) \circ \eta_{(V,W)}. $$ Note that: $$ \begin{array}{rccl} \Hom(f,g): & \Hom(V,W) & \to & \Hom(V',W'), \\ \Phi(f,g): & \Phi(V,W) & \to & \Phi(V',W'), \\ \eta_{(V,W)}: & \Hom(V,W) & \to & \Phi(V,W), \\ \eta_{(V',W')}: & \Hom(V',W') & \to & \Phi(V',W'), \end{array} $$ in other words all of the above compositions are defined, and both the morphism $\eta_{(V',W')} \circ \Hom(f,g)$ and the morphism $\Phi(f,g)\circ \eta_{(V,W)}$ map $\Hom(V,W) \to \Phi(V',W')$.

We now propose the following formula for the natural transformation $\eta$, which, given an object $(V,W)$ in the source category $(\Vec^{op} \times \Vec)$ returns a morphism between the two corresponding objects in the target category $(\Vec)$ defined by the functors $\Hom$ and $\Phi$, namely a morphism $\Hom(V,W) \to \Phi(V,W)$. In other words: $$ \eta: (V,W) \mapsto [ \Hom(V,W) \to \Phi(V,W) ] . $$ Anyway, we have that $\eta_{(V,W)}$ maps between the spaces $\Hom(V,W)$ and $\Phi(V,W) = \Hom(V \otimes W^*, \mathbb{R})$ according to the formula: $$ \sigma \mapsto [ \eta_{(V,W)}(\sigma): \sum_{i,j} v_i \otimes \psi_j \mapsto \sum_{i,j} \psi_j(\sigma(v_i)) ] , $$ i.e. $\sigma: V \to W$ and $\eta_{(V,W)}(\sigma): V \otimes W^* \to \mathbb{R}$, both linear.

So anyway, showing that $\eta$ is a natural transformation again comes down to verifying the equality (for arbitrary $(V,W) \overset{(f,g)}{\to} (V',W')$): $$ \eta_{(V',W')} \circ \Hom(f,g) = \Phi(f,g) \circ \eta_{(V,W)}. $$ Remember that both sides map $\Hom(V,W) \to \Phi(V',W')$, so if we show that both sides evaluate to the same value when applied to an arbitrary $\sigma \in \Hom(V,W)$, we will have shown the equality between functions for arbitrary $(V,W) \overset{(f,g)}{\to} (V',W')$ and thus that $\eta$ is a natural transformation, i.e. all we have to do is show, for arbitrary $\sigma \in \Hom(V,W)$: $$ ( \eta_{(V',W')} \circ \Hom(f,g) ) (\sigma) = (\Phi(f,g) \circ \eta_{(V,W)})(\sigma). $$ Let's calculate the left-hand side first: $$ (\eta_{(V',W')} \circ \Hom(f,g))(\sigma) = \eta_{(V',W')}( \Hom(f,g)(\sigma) ) . $$ Now by definition of $\Hom(f,g): \Hom(V,W) \to \Hom(V',W')$, we have that: $$ \Hom(f,g)(\sigma) = g \circ \sigma \circ f. $$ Since $f: V' \to V, \sigma: V \to W,$ and $g: W \to W'$, we have that $g \circ \sigma \circ f: V' \to V \to W \to W'$ i.e. $g \circ \sigma \circ f: V' \to W'$, such that: $$\Hom(f,g)(\sigma)=g \circ \sigma \circ f \in \Hom(V',W'), $$ as expected and required. Also, since $\eta_{(V',W')}: \Hom(V',W') \to \Phi(V',W')$, we have as a result that $\eta_{(V',W')}(g \circ \sigma \circ f)$ is well-defined. Anyway, we have by definition of $\eta_{(V',W')}$ that: $$ \eta_{(V',W')}(g \circ \sigma \circ f) = \left[\sum_{i,j} v'_i \otimes \psi'_j \mapsto \sum_{i,j} \psi'_j((g \circ \sigma \circ f) (v'_i) ) = \sum_{i,j} \psi'_j(g(\sigma (f(v'_i)))) \right]. $$ Now let's calculate the right-hand side: $$ (\Phi(f,g) \circ \eta_{(V,W)})(\sigma) = \Phi(f,g) ( \eta_{(V,W)}(\sigma).$$ Now we have again by definition of $\eta$ that:$$ \eta_{(V,W)}(\sigma) = \left[ \sum_{i,j} v_i \otimes \psi_j \mapsto \sum_{i,j} \psi_j(\sigma(v_i)) \right] . $$ Allow us to recall the definition of the morphism part of $\Phi$: $$ \Phi(f,g): \tau \mapsto \tau \circ (f \otimes g^*). $$ Thus, we can now evaluate the rest of the expression on the right-hand side: $$ \Phi(f,g) \left( \left[ \sum_{i,j} v_i \otimes \psi_j \mapsto \sum_{i,j} \psi_j(\sigma(v_i)) \right] \right) = \left[ \sum_{i,j} v_i \otimes \psi_j \mapsto \sum_{i,j} \psi_j(\sigma(v_i)) \right] \circ (f \otimes g^*) $$ Allow us to pause for a moment to recall the definition of $(f \otimes g^*): V' \otimes (W')^* \to V \otimes W^*$: $$ (f \otimes g^*): v' \otimes \psi' \mapsto (f(v')) \otimes (\psi' \circ g). $$ Hence, the above equals: $$ = \left[ \sum_{i,j} v'_i \otimes \psi'_j \mapsto \sum_{i,j} (f(v'_i)) \otimes (\psi'_j \circ g) \mapsto \sum_{i,j} (\psi'_j \circ g)(\sigma (f(v_i'))) \right] $$ $$ = \left[ \sum_{i,j} v_i' \otimes \psi_j' \mapsto \sum_{i,j} \psi'_j(g(\sigma(f(v'_i)))) \right]. $$ Hence, we have shown that: $$ \eta_{(V',W')} (\Hom(f,g)(\sigma) = \left[ \sum_{i,j} v_i' \otimes \psi_j' \mapsto \sum_{i,j} \psi_j'(g(\sigma(f(v_i')))) \right] = \Phi(f,g)(\eta_{(V,W)}(\sigma) ). $$ Since $(V,W) \overset{(f,g)}{\to}(V',W')$ and $\sigma \in \Hom(V,W)$ were arbitrary, this completes the proof that $\eta$ is in fact a natural transformation from $\Hom(-,-)$ to $\Phi(-,-)$.

Claim: There is a natural transformation from $\Phi(-,-)$ to $\Hom(-,-)$ which is inverse to $\eta$, denote it tentatively $\eta^{-1}$.

In this claim, there are really two claims to unpack: (1) first, that there exists a natural transformation $\eta^{-1}$ taking $\Phi(-,-)$ to $\Hom(-,-)$, and (2) that $\eta(\eta^{-1}(\Phi(-,-)))=\Phi(-,-)$ and that $\eta^{-1}(\eta(\Hom(-,-)))=\Hom(-,-)$. (So perhaps really more like three claims packed into one.)

(1) We will start by trying to find a formula for $\eta^{-1}$ and then evaluating the first (sub-)claim.

We would need $\eta^{-1}$ to satisfy: $$ \eta^{-1}_{(V',W')}\circ \Phi(f,g) = \Hom(f,g) \circ \eta^{-1}_{(V,W)} $$ for arbitrary $(V,W),(V',W') \in \Vec^{op}\times \Vec, f: V' \to V, g: W \to W'$ linear.

Note that: $$ \begin{array}{rccl} \Phi(f,g): & \Phi(V,W) & \to & \Phi(V',W'), \\ \eta^{-1}_{(V',W')}: & \Phi(V',W') & \to & \Hom(V',W'), \\ \eta^{-1}_{(V,W)}: & \Phi(V,W) & \to & \Hom(V,W), \\ \Hom(f,g): & \Hom(V,W) & \to & \Hom(V',W'), \end{array} $$ hence both sides of the above equation map $\Phi(V,W) \to \Hom(V',W')$.

Let us start with $\tau \in \Phi(V,W) = \Hom(V \otimes W^*, \mathbb{R})$. Then $\tau$ is a linear map taking tensors of the form $\sum_{i,j} v_i \otimes \psi_j$, with $v_i \in V$ and $\psi_j \in W^*$, to real numbers. Applying $\eta^{-1}_{(V,W)}$ to $\tau$, we somehow want to get out of it a map in $\Hom(V,W)$, i.e. a linear transformation from $V$ to $W$.

Since $W$ is finite-dimensional, we can fix for it a basis $(w_1, \dots, w_n)$. Corresponding to that let $(\psi_1, \dots, \psi_n)$ be the dual basis of $W^*$. Then, given $\tau \in \Hom(V \otimes W^*, \mathbb{R})=\Phi(V,W)$, we will define $\eta^{-1}_{(V,W)}$ by: $$ \eta^{-1}_{(V,W)}: \tau \mapsto \left[ v \mapsto \sum_{i=1}^{n} \tau(v \otimes \psi_i) \cdot w_i \right] . $$ Now we have to begin the process of checking that this is a natural transformation, i.e. that for arbitrary $\tau \in \Hom(V \otimes W^*, \mathbb{R})= \Phi(V,W)$, that: $$ \eta^{-1}_{(V',W')}(\Phi(f,g)(\tau)) = \Hom(f,g) (\eta^{-1}_{(V,W)}(\tau) ). $$

As before, let's start with the left-hand side first. We have that: $$ \Phi(f,g)(\tau) = \tau \circ (f \otimes g^*). $$ From this it follows that: $$ \eta^{-1}_{(V',W')}(\Phi(f,g)(\tau) )= \eta^{-1}_{(V',W')}(\tau \circ (f \otimes g^*)) = \left[ v' \mapsto \sum_{i=1}^{n} (\tau \circ (f \otimes g^*) )(v' \otimes \psi_i') \cdot w_i' \right] $$ $$ = \left[ v' \mapsto \sum_{i=1}^{n} \tau( (f \otimes g^*)(v' \otimes \psi_i') ) \cdot w_i' = \sum_{i=1}^{n} \tau( f(v') \otimes (\psi_i' \circ g) ) \cdot w_i' \right]. $$ Note that $\psi_i' \circ g: W \to W' \to \mathbb{R}$, i.e. $\psi_i' \circ g \in W^*$, and $f: V' \to V$, so $f(v') \in V$, such that $f(v') \otimes (\psi_i' \circ g ) \in V \otimes W^*$, and thus $\tau ( f(v') \otimes (\psi_i' \circ g) )$ is well-defined. Hence we have: $$ \eta^{-1}_{(V',W')}(\Phi(f,g) (\tau) ) = \left[ v' \mapsto \sum_{i=1}^{n} \tau ( f(v') \otimes (\psi_i'\circ g) ) \cdot w_i' \right], $$ which is clearly a map in $\Hom(V',W')$, as expected and required.

We will now calculate the right-hand side, as follows; we start with: $$\eta^{-1}_{(V,W)}(\tau) = \left[ v \mapsto \sum_{i=1}^{n} \tau(v \otimes \psi_i)\cdot w_i \right], $$ which is simply the definition of $\eta^{-1}_{(V,W)}$ as stated above. Clearly, this is in $\Hom(V,W)$ as expected and required. Now we apply the morphism $\Hom(f,g)$ to get an element in $\Hom(V',W')$, hopefully the same as before: $$\Hom(f,g)(\eta^{-1}_{(V,W)}(\tau) ) = \Hom(f,g)\left( \left[ v \mapsto \sum_{i=1}^{n} \tau(v \otimes \psi_i) \cdot w_i \right] \right) = g \circ \left[ v \mapsto \sum_{i=1}^{n} \tau(v \otimes \psi_i) \cdot w_i \right] \circ f. $$ Remembering that $f:V' \to V, f: v' \mapsto v$, we can rewrite the above as: $$ = g \circ \left[ v' \mapsto f(v') \mapsto \sum_{i=1}^{n} \tau( f(v') \otimes \psi_i ) \cdot w_i \right] = g \circ \left[ v' \mapsto \sum_{i=1}^{n} \tau( f(v') \otimes \psi_i ) \cdot w_i \right].$$ Remember now that $g: W \to W'$ is a linear transformation, hence we can distribute it over the sum to get: $$ = \left[ v' \mapsto g \left( \sum_{i=1}^{n} \tau( f(v') \otimes \psi_i ) \cdot w_i \right) \right] = \left[ v' \mapsto \sum_{i=1}^{n} \tau(f(v') \otimes \psi_i )\cdot g(w_i) \right]. $$ Remember, that since $f(v') \in V$, and $\tau: V \otimes W^* \to \mathbb{R}$ that $\tau (f(v') \otimes \psi_i) \in \mathbb{R}$, which is why $g$ distributes over it. Now let's remember that $(\psi_1, \dots, \psi_n)$ is the dual basis to $(w_1, \dots, w_n)$, and $(\psi_1',\dots, \psi_n')$ is the dual basis to $(w_1', \dots, w_n')$. Thus, if we want to claim that the expression we have just derived is equal to the expression for $\eta^{-1}_{(V',W')}(\Phi(f,g) (\tau) )$, then we must show that: $$ w_i' = g(w_i) \implies \psi_i = \psi_i' \circ g,$$ and $$\psi_i' \circ g = \psi_i \implies w_i' = g(w_i).$$

Let's start with the first direction. Assume that $w_i'=g(w_i)$. Then: $$1=\psi_i'(w_i')=\psi_i'(g(w_i)) = (\psi_i' \circ g)(w_i).$$ Since $\psi_i$ is the unique element of $W^*$ such that $\psi_i(w_i)=1$, and we have shown above that $(\psi_i' \circ g)(w_i)=1$, it follows that $\psi_i = \psi_i' \circ g$, as we wanted to show.

Now let us show the second direction. Assume that $\psi_i' \circ g = \psi_i$. Then: $$1 = \psi_i(w_i) = (\psi_i' \circ g)(w_i) = \psi_i'(g(w_i)).$$ Since $w_i'$ is the unique element of $W'$ such that $\psi_i'(w_i')=1$, and we have shown above that $\psi_i' (g(w_i))=1$, it follows that $g(w_i)=w_i'$, which is what we wanted to show.

Therefore, we can in full faith and confidence interchange $g(w_i)$ with $w_i'$ whenever we interchange $\psi_i$ with $\psi_i' \circ g$. In particular, we have shown that: $$\left[ v' \mapsto \sum_{i=1}^{n} \tau(f(v') \otimes \psi_i ) \cdot g(w_i) \right] = \left[ v' \mapsto \sum_{i=1}^{n} \tau( f(v') \otimes (\psi_i' \circ g) ) \cdot w_i' \right]. $$

In conclusion, we have shown that: $$\eta^{-1}_{(V',W')}(\Phi(f,g)(\tau) ) = \left[ v' \mapsto \sum_{i=1}^{n} \tau( f(v') \otimes (\psi'_i \otimes g) )\cdot w_i' \right] = \Hom(f,g)(\eta^{-1}_{(V,W)}(\tau) ) .$$ Thus $\eta^{-1}$ really is a natural transformation, as had previously been claimed.

(2) Now, given that there is a natural transformation $\eta^{-1}$ taking $\Phi(-,-)$ to $\Hom(-,-)$, we want to see whether this natural transformation is inverse to the natural transformation $\eta$ we had defined earlier, which takes $\Hom(-,-)$ to $\Phi(-,-)$.

In particular, if we want to say that $\eta^{-1}$ is inverse to $\eta$, then we want to show that $\eta^{-1} \circ \eta$ is the identity natural transformation $\Hom(-,-) \to \Hom(-,-)$, and that $\eta \circ \eta^{-1}$ is the identity natural transformation $\Phi(-,-) \to \Phi(-,-)$. Thus we have to show that: $$\Hom(f,g) \circ \eta_{(V,W)}^{-1} \circ \eta_{(V,W)} = \eta^{-1}_{(V',W')} \circ \eta_{(V',W')} \circ \Hom(f,g), $$ and $$ \eta_{(V',W')} \circ \eta^{-1}_{(V',W')} \circ \Phi(f,g) = \Phi(f,g) \circ \eta_{(V,W)} \circ \eta^{-1}_{(V,W)}.$$ Note that this will follow if we can show that: $$\begin{array}{rcl} \eta_{(V,W)}^{-1} \circ \eta_{(V,W)} & = & \id_{\Hom(V,W)}, \\ \eta^{-1}_{(V',W')} \circ \eta_{(V',W')} & = & \id_{\Hom(V',W')}, \\ \eta_{(V,W)} \circ \eta^{-1}_{(V,W)} & = & \id_{\Phi(V,W)}, \\ \eta_{(V',W')} \circ \eta^{-1}_{(V',W')} & = & \id_{\Phi(V',W')}, \end{array}$$ since the identity morphisms satisfy the required commutativity relations by definition, i.e. by definition of identity morphism it holds that: $$\begin{array}{rcl} \Hom(f,g) \circ \id_{\Hom(V,W)} & = & \id_{\Hom(V',W')} \circ \Hom(f,g) \\ \id_{\Phi(V',W')} \circ \Phi(f,g) & = & \Phi(f,g) \circ \id_{\Phi(V,W)}. \end{array} $$ Note also that if we show for arbitrary $(\tilde{V}, \tilde{W}) \in \Vec^{op} \times \Vec$ that $$\begin{array}{rcl} \eta^{-1}_{(\tilde{V}, \tilde{W})} \circ \eta_{(\tilde{V}, \tilde{W}) } & = & \id_{\Hom(\tilde{V}, \tilde{W} )}, \\ \eta_{ (\tilde{V}, \tilde{W} ) } \circ \eta^{-1}_{ (\tilde{V}, \tilde{W} ) } & = & \id_{ \Phi( \tilde{V}, \tilde{W} ) }, \end{array} $$ then we will have proven all four of the above claims (since again $(\tilde{V}, \tilde{W})$ is chosen arbitrarily, and thus can stand for either pair $(V,W)$ or $(V',W')$ if we like). Showing the first equation: $$\eta^{-1}_{(\tilde{V}, \tilde{W})} \circ \eta_{(\tilde{V}, \tilde{W})} = \id_{\Hom(\tilde{V}, \tilde{W})}, $$ will prove the first claim, that $\eta^{-1} \circ \eta$ is the identity natural transformation $\Hom(-,-) \to \Hom(-,-)$. Showing the second equation: $$\eta_{(\tilde{V}, \tilde{W})} \circ \eta^{-1}_{(\tilde{V}, \tilde{W})} = \id_{\Phi(\tilde{V}, \tilde{W})}, $$ will prove the second claim, that $\eta \circ \eta^{-1}$ is the identity natural transformation $\Phi(-,-) \to \Phi(-,-)$.

Let $\sigma \in \Hom(\tilde{V}, \tilde{W})$ be arbitrary, we will now prove the first claim by showing that: $$\eta^{-1}_{(\tilde{V}, \tilde{W})}( \eta_{(\tilde{V}, \tilde{W})} (\sigma) ) = \sigma. $$ By definition of $\eta$ given above, we have that: $$ \eta_{(\tilde{V}, \tilde{W})}(\sigma) = \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \tilde{\psi}_j(\sigma(\tilde{v}_i) ) \right]. $$ Then it follows that: $$\eta^{-1}_{(\tilde{V}, \tilde{W})} (\eta_{(\tilde{V}, \tilde{W})} (\sigma )) = \eta^{-1}_{(\tilde{V}, \tilde{W})} \left( \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \tilde{\psi}_j ( \sigma (\tilde{v}_i) ) \right] \right), $$ using the definition of $\eta^{-1}$ given above: $$ = \left[ \tilde{v} \mapsto \sum_{k=1}^{n} \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \tilde{\psi}_j ( \sigma (\tilde{v}_i) ) \right] (\tilde{v} \otimes \tilde{\psi}_k) \cdot \tilde{w}_k \right] = \left[ \tilde{v} \mapsto \sum_{k=1}^{n} (\tilde{\psi}_k(\sigma(\tilde{v})))\cdot \tilde{w}_k \right]. $$ So we are done as long as we can show that: $$\sigma(\tilde{v}) = \sum_{k=1}^{n} (\tilde{\psi}_k(\sigma(\tilde{v}))) \cdot \tilde{w}_k .$$ Remember that $\tilde{\psi}_{k}(\tilde{w}_{\ell})=\delta_{k,\ell}$ by definition of these quantities (where $\delta_{k,\ell}$ is the Kronecker delta). Since $(w_1, \dots, w_n)$ is a basis for $W$, we can write uniquely: $$ \sigma(\tilde{v}) = \sum_{\ell=1}^{n} a_{\ell} \cdot \tilde{w}_{\ell}, $$ for some yet to be determined quantities $a_{\ell} \in \mathbb{R}$. Then for each $1 \le k \le n$, using linearity of $\tilde{\psi}_k$: $$\tilde{\psi}_{k} (\sigma(\tilde{v})) = \tilde{\psi}_{k} \left( \sum_{\ell=1}^{n} a_{\ell} \cdot \tilde{w}_{\ell} \right) = \sum_{\ell=1}^{n} a_{\ell} \tilde{\psi}_{k}(\tilde{w}_{\ell}) = \sum_{\ell=1}^{n} a_{\ell} \delta_{k, \ell} = a_k. $$ Therefore, we have shown that: $$\sigma(\tilde{v}) = \sum_{\ell=1}^{n} a_{\ell} \cdot \tilde{w}_{\ell} = \sum_{k=1}^{n} a_k \cdot \tilde{w}_k = \sum_{k=1}^{n} (\tilde{\psi}_k(\sigma(\tilde{v})) )\cdot \tilde{w}_k, $$ as we had wanted to do. Then for arbitrary $\tilde{v} \in \tilde{V}$ and arbitrary $\sigma \in \Hom(\tilde{V}, \tilde{W})$: $$\eta^{-1}_{(\tilde{V},\tilde{W})} (\eta_{(\tilde{V}, \tilde{W})} (\sigma(\tilde{v})) ) = \sum_{k=1}^{n} (\tilde{\psi}_k(\sigma(\tilde{v})))\cdot \tilde{w}_k = \sigma(\tilde{v}) , $$ $$ \iff \eta^{-1}_{(\tilde{V}, \tilde{W})} ( \eta_{(\tilde{V}, \tilde{W})} (\sigma)) = \sigma \iff \eta^{-1}_{(\tilde{V}, \tilde{W})} \circ \eta_{(\tilde{V}, \tilde{W})} = \id_{\Hom(\tilde{V}, \tilde{W})}.$$ It follows that $\eta^{-1} \circ \eta$ is the identity natural transformation $\Hom(-,-) \to \Hom(-,-)$.

Now let $\tau \in \Phi(\tilde{V}, \tilde{W}) = \Hom(\tilde{V} \otimes (\tilde{W})^*, \mathbb{R})$ be arbitrary. We prove the second claim by showing: $$\eta_{(\tilde{V}, \tilde{W})} (\eta^{-1}_{(\tilde{V}, \tilde{W})}(\tau) ) = \tau. $$ One has: $$\eta^{-1}_{(\tilde{V}, \tilde{W})}(\tau) = \left[ \tilde{v} \mapsto \sum_{k=1}^{n} \tau(\tilde{v} \otimes \tilde{\psi}_k) \cdot \tilde{w}_k \right], $$ and thus further that: $$ \eta_{(\tilde{V}, \tilde{W})} (\eta^{-1}_{(\tilde{V}, \tilde{W})}(\tau) ) = \eta_{(\tilde{V}, \tilde{W})}\left( \left[ \tilde{v} \mapsto \sum_{k=1}^{n} \tau(\tilde{v} \otimes \tilde{\psi}_k) \cdot \tilde{w}_k \right] \right) = \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \tilde{\psi}_j \left( \left[ \tilde{v} \mapsto \sum_{k=1}^{n} \tau(\tilde{v} \otimes \tilde{\psi}_k) \cdot \tilde{w}_k ) \right] (\tilde{v}_i) \right) \right] $$ $$ = \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \tilde{\psi}_j \left( \sum_{k=1}^{n} \tau(\tilde{v}_i \otimes \tilde{\psi}_k) \cdot \tilde{w}_k \right) \right] = \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \sum_{k=1}^{n} \tau(\tilde{v}_i \otimes \tilde{\psi}_k ) \tilde{\psi}_j (\tilde{w}_k) \right] $$ $$\left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \sum_{k=1}^{n}\tau(\tilde{v}_i \otimes \tilde{\psi}_k) \delta_{jk} \right] = \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \tau(\tilde{v}_i\otimes \tilde{\psi}_j ) \right] \in \Hom(\tilde{V} \otimes (\tilde{W})^*, \mathbb{R} )=\Phi(\tilde{V}, \tilde{W}). $$ Now by linearity of $\tau$ we actually have immediately that: $$\sum_{i,j} \tau(\tilde{v}_i \otimes \tilde{\psi}_j) = \tau\left( \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \right). $$ To be even more explicit (the first equality is a definition, the second holds by linearity): $$\tau = \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \tau \left( \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \right) \right] = \left[ \sum_{i,j} \tilde{v}_i \otimes \tilde{\psi}_j \mapsto \sum_{i,j} \tau(\tilde{v}_i \otimes\tilde{\psi}_j) \right]. $$ In conclusion, we have shown that, for arbitrary $\tau \in \Phi(\tilde{V}, \tilde{W})$ that: $$\eta_{(\tilde{V},\tilde{W})}( \eta^{-1}_{(\tilde{V}, \tilde{W})} (\tau) ) = \tau \iff \eta_{ (\tilde{V}, \tilde{W} )}\circ \eta^{-1}_{(\tilde{V}, \tilde{W})} = \id_{\Phi(\tilde{V}, \tilde{W})}.$$ It follows that $\eta \circ \eta^{-1}$ is the identity natural transformation $\Phi(-,-) \to \Phi(-,-)$.

Thus not only (1) is $\eta^{-1}$ a natural transformation, but finally (2) we have shown that it is indeed inverse to the natural transformation $\eta$. Thus we have established the existence of a \textbf{natural isomorphism} between the functors $\Hom(-,-)$ and $\Phi(-,-)$.

In particular, we have shown that, for all $(V,W) \in \Vec^{op} \times \Vec$, we have: $$\Hom(V,W) \cong \Hom(V\otimes W^*, \mathbb{R} ), $$ in the category of finite-dimensional vector spaces, i.e. that $\Hom(V,W)$ and $\Hom(V\otimes W^*, \mathbb{R})$ are isomorphic as vector spaces for any choice of two finite-dimensional vector spaces $(V,W)$.

Answer 3

$\newcommand{\Vec}{\mathsf{Vec}}$$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\id}{\operatorname{id}}$My answer exceeded the character limit, so here's the last part:

Finally, to get the general case, consider the following functor: $$\bigotimes: \Vec^{\times k} \to \Vec, $$ which acts by: $$\bigotimes(V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k. $$

Claim: $\bigotimes$ is actually a functor.

We already gave the proposed object part above ($\bigotimes(V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k$), now the morphism part is proposed to be the following: $$\bigotimes(f_1, \dots, f_k) = f_1 \otimes \dots \otimes f_k, $$ perhaps unsurprisingly. Now we need to check that this actually constitutes a functor.

Let $\id_{(V_1,\dots,V_k)}=(\id_{V_1}, \dots, \id_{V_k})$ (this holds by definition of product category). Then: $$\bigotimes \id_{(V_1, \dots, V_k)} = \bigotimes (\id_{V_1}, \dots, \id_{V_k}) = \id_{V_1} \otimes \dots \otimes \id_{V_k}. $$ For $\bigotimes$ to be a functor, it is necessary that this be the identity morphism on $\bigotimes (V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k$. However, by definition of the tensor product, it is actually relatively clear that: $$\id_{V_1} \otimes \dots \otimes \id_{V_k}=\id_{V_1 \otimes \dots \otimes V_k} = \id_{\bigotimes(V_1, \dots, V_k)} . $$ Thus $\bigotimes$ preserves identity morphisms, $$\bigotimes \id_{(V_1, \dots, V_k)} = \id_{\bigotimes (V_1, \dots, V_k)}, $$ as required. Now we need to show that $\bigotimes$ is compatible with composition of morphisms, i.e. given: $$(V_1, \dots, V_k) \overset{(f_1, \dots, f_k)}{\to} (V_1', \dots, V_k') \quad \text{and} \quad (V_1',\dots, V_k') \overset{(g_1, \dots, g_k)}{\to} (V_1'', \dots, V_k''), $$ one has that: $$\bigotimes\left( (g_1, \dots, g_k) \circ (f_1, \dots, f_k) \right) = \bigotimes(g_1, \dots, g_k) \circ \bigotimes (f_1, \dots, f_k). $$ Evaluating the left-hand side first, we have: $$\bigotimes \left( (g_1, \dots, g_k) \circ (f_1, \dots, f_k) \right) = \bigotimes (g_1\circ f_1, \dots, g_k \circ f_k)= (g_1 \circ f_1) \otimes \dots \otimes (g_k \circ f_k). $$ Now evaluating the right-hand side, one finds: $$\bigotimes (g_1, \dots, g_k) \circ \bigotimes (f_1, \dots, f_k) = (g_1 \otimes \dots \otimes g_k) \circ (f_1 \otimes \dots \otimes f_k). $$ We showed, when proving that $\Phi$ is a functor, that the tensor product $\otimes$ plays nicely with composition $\circ$, so we have finally that: $$(g_1 \otimes \dots \otimes g_k) \circ (f_1 \otimes \dots \otimes f_k) = (g_1 \circ f_1) \otimes \dots \otimes (g_k \circ f_k). $$ In conclusion, we have shown that: $$\bigotimes\left( (g_1, \dots, g_k) \circ (f_1, \dots, f_k) \right) = (g_1 \circ f_1) \otimes \dots \otimes (g_k\circ f_k) = \bigotimes(g_1, \dots,g_k ) \circ \bigotimes(f_1, \dots, f_k), $$ hence $\bigotimes$ is compatible with composition of morphisms and thus really is a functor, as claimed.

Claim: $op: \Vec \to \Vec^{op}$ is a contravariant functor.

The object part is the same as the object part of the identity functor: $op: V \mapsto V$.

For the morphism part, we have the following rule (as follows from the definition of dual category): $$op: \left( V \overset{f}{\to} V' \right) \mapsto \left( V \overset{f^{op}}{\gets} V' \right). $$ Note that while $f$ is always a function (a linear transformation in fact), in general $f^{op}$ is not, i.e. only a morphism but not a function (although one can define it as a function in the case that $f$ happens to be an isomorphism).

Anyway, we need to show that $op$ is compatible with identity morphisms: $op(\id_V) = \id_{op(V)}=\id_V$. One has that: $$op \left( V \overset{id_V}{\to} \id_V \right) \mapsto \left( V \overset{\id_V}{\gets} V \right). $$ Now since $\id_V$ is an isomorphism, we have that the left and right hand sides are equal, hence $op(\id_V)= \id_V$, as required.

Now we show that $op$ is compatible with composition of morphisms in a contravariant manner. $$op(g \circ f) = \left( V \overset{(g\circ f)^{op}}{\gets} V'' \right) = \left( V \overset{f^{op}}{\gets} V' \right) \circ \left( V' \overset{g^{op}}{\gets} V'' \right) = op(f) \circ op(g), $$ as expected and required.

Because we have a natural isomorphism between $\Hom(-,-)$ and $\Phi(-,-)$, it follows that we also have a natural isomorphism between $\Hom((\bigotimes(-,\dots, -))^{op}, - )$ and $\Phi((\bigotimes(-,\dots, -))^{op},- )$, i.e. for any $k-$tuple of finite-dimensional vector spaces $(V_1, \dots, V_k)$, and any finite-dimensional vector space $W$, one has that: $$\Hom(V_1 \otimes \dots \otimes V_k, W) \cong \Hom(V_1 \otimes \dots \otimes V_k \otimes W^*, \mathbb{R} ), $$ the isomorphism being natural.

Claim: $\times k: \Vec \to \Vec^{\times k}$ is a functor.

For the object part, we have simply: $$\times k (V) = \underbrace{(V, \dots, V)}_{k\text{ times}}, $$ and for the morphism part, given $V \overset{f}{\to} V'$, one has: $$\times k(f) = \underbrace{(V, \dots, V)}_{k\text{ times}} \overset{\underbrace{(f, \dots, f)}_{k \text{ times}}}{\to} \underbrace{(V', \dots, V')}_{k\text{ times}}.$$ Now one has fairly immediately that: $$\times k(\id_V) = (\id_V, \dots, \id_V) = \id_{(V,\dots, V)}. $$ Likewise, one also has fairly immediately that: $$\times k (g \circ f) = (g\circ f, \dots, g\circ f) = (g, \dots, g)\circ(f,\dots,f) = \times k(g) \circ \times k(f). $$ Thus $\times k: \Vec \to \Vec^{\times k}$ is a functor, as claimed.

Thus composing functors again, we get a natural isomorphism between $$\Hom\left( \left(\bigotimes( \times k (-) )\right)^{op}, - \right) \quad \text{and} \quad \Phi\left( \left(\bigotimes(\times k (-)) \right)^{op},- \right),$$ thus, as we wanted to show, for any choice of two finite-dimensional vector spaces $V$ and $W$, there is a natural isomorphism between: $$\Hom(\underbrace{V \otimes \dots \otimes V}_{k \text{ times}}, W) \quad \text{and} \quad \Hom(\underbrace{V \otimes \dots \otimes V}_{k\text{ times}} \otimes W^*, \mathbb{R}). $$