Prove naturality of unit and counit in Tensor-hom adjunction

Question

I am learning some basics from Category Theory and I am currently working on understanding Tensor-hom adjunction. I was reading a Wiki article about it (https://en.wikipedia.org/wiki/Tensor-hom_adjunction), but I am confused about the following:

When we construct a unit and counit as there as following: The unit

$$\varepsilon : FG \to 1_{\mathcal C}$$ has components $$\varepsilon_Z : Hom_S (X,Z) \otimes_R X \to Z$$ given by evaluation: \begin{align*} & \phi \in Hom_S(X,Z) \quad \textrm{i} \quad x\in X \\ & \varepsilon(\phi \otimes x) = \phi(x). \end{align*} And counit with components \begin{align*} & \eta: 1_{\mathcal D} \to GF \\ & \eta_Y : Y \to Hom_S(X, Y \otimes_R X) \end{align*} we define as $$\eta_Y(y)(t) = y \otimes t \quad \textrm{for} \quad t \in X.$$ Counit and unit equations can now be easily checked. What about naturality? How to formally prove that these transformations are indeed natural? Also, I was wondering if it is true that evaluation is always a natural transformation.

My work on proving $\varepsilon$ is natural: To prove this, we have to show the following diagram commutes:

But, I am not sure what exactly $FG(f)$ is (it's defined for objects, but what about arrows?) and how to prove that this commutes.

Edit: So, we have functors defined as \begin{align*} F(Y) &= Y \otimes_R X \quad \textrm{za} \quad Y \in \mathcal D \\ G(Z) &= Hom_S(X,Z) \quad \textrm{za} \quad Z \in \mathcal C \end{align*} And what we are looking for is $FG(f)$ where $f: Z \to Y$. In the definition of functor, we have that if $f: X \to Y$ then $F(f): F(X) \to F(Y)$. So it is not hard to conclude that $$G(f) : Hom_S (X,Z) \to Hom_S(X,Y)$$ and $$FG(f) : Hom_S(X,Z) \otimes_R X \to Hom_S(X,Y) \otimes_R X$$ Is this enough to prove that such a diagram commutes?

Answer 1

To put our discussion in the comments into an answer. $\newcommand{\Hom}{\operatorname{Hom}}$

First of all, let's write down how these functors act on morphisms. Let $f\colon A\to B$. Then, $\Hom(X,f)\colon \Hom(X,A)\to \Hom(X,B)$. So, we take some morphism $s\colon X\to A$ and need to produce morphism $\Hom(X,f)(s)\colon X\to B$ which we can do easily with composition $X\stackrel{s}{\to} A \stackrel{f}{\to} B$, i.e. $\Hom(X,f)(s) = f\circ s$.

For tensor product, note that given any two linear maps $f\colon A\to A'$ and $g\colon B\to B'$, map $h\colon A\times B\to A'\otimes B'$ defined by $h(a,b) = f(a)\otimes g(b)$ is bilinear, so by the universal property there is a unique linear map $f\otimes g\colon A\otimes B\to A'\otimes B'$ such that $(f\otimes g)(a\otimes b) = f(a)\otimes g(b)$. So, given $f\colon A\to B$, we need $f\otimes X\colon A\otimes X\to B\otimes X$ which we can define simply as $f\otimes\mathrm{id}_X$.

We are now ready to check naturality:

\begin{array}{cccc} \Hom(X,A)&\!\!\!\!\otimes&\!\!\!\! X &\stackrel{\epsilon_A}{\longrightarrow} &A &\stackrel{f}{\longrightarrow} &B\\ \qquad\qquad s&\!\!\!\!\otimes&\!\!\!\! x &{\mapsto} &s(x) &{\mapsto} & f(s(x)) \end{array}

\begin{array}{cccc} \Hom(X,A)&\!\!\!\!\otimes&\!\!\!\! X &\stackrel{(f\circ-)\otimes\mathrm{id}_X}{\longrightarrow} &\Hom(X,B)&\!\!\!\!\otimes&\!\!\!\! X &\stackrel{\epsilon_B}{\longrightarrow} &B\\ \qquad\qquad s&\!\!\!\!\otimes&\!\!\!\! x &{\mapsto} &\qquad(f\circ s)&\!\!\!\!\otimes&\!\!\!\! x &{\mapsto} & (f\circ s)(x) \end{array}

which are obviously equal.

Answer 2

Here is a great theorem:

Assume two categories $C$ and $D$ and a functor $G : D \to C$. Now, assume for each $c \in C$, we have a universal arrow $(d , \eta_d : c \to G d)$ from $c$ to $G$. Then we can construct a functor $F : C \to D$ such that $F \dashv G$.

Proof sketch: we define $F c$ to be the $d$ such that $ (d , \eta_c : c \to G d)$ is the universal arrow. For $f : c \to c;'$, we can define the action an morphisms as follows. We have a diagram:

Here, the dotted line is guarenteed to exist uniquely via the fact that $\eta_c$ is a universal arrow. We define $F(f)$ to be this dotted morphism. We of course must verify that the assignment we have defined actually assembles into a functor $F : C \to D$. I leave this to you. It is now immideate that $\eta : 1_C \Rightarrow G \circ F$. There is more to check to ensure that we have an adjunction. I leave this to you. $\square$

You can find this theorem in 4.6 of Riehl's category theory in context.

Now, lets use this theorem to solve your problem. You already have a functor $\hom(X , -)$. So we just need to prove that, for each vector space $W$, we have a universal arrow $W \to \hom (X , V)$. Further, since we want to show that $X \otimes -$ is the left adjoint, we need to have $V \equiv X \otimes W$.

You have already defined the universal arrow $\eta_W$. You just need to show that it is actually universal. Do you think you can take it from here? If not, I'll edit this answer to complete the proof.