Solution check request for an exercise in Arbib and Manes's showing the existence of $\textbf{equalizer}$ in the category of vector spaces

Question

The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes

Let $f,g:X\rightarrow Y$ be linear maps between vector spaces $X,Y$. Using the following exercise (one in the block quote), show that $\text{eq}(f,g)$ exists, where $\text{eq}(f,g)$ denotes the equalizer of $f,g$. There is also another question with the same wording for coequalizer of $(f,g)$, but I will make a separate post for it.

Let $X$ be a vector space. If $A$ is a subset of $X$, show that there is at most one vector space structure on $A$ such that the inclusion map $i:A\rightarrow X$ is linear, and that this occurs iff $A$ is a subspace of $X$. If $E$ is an equivalence relation on $X,$ show that there is at most one vector space structure on $X/E$ such that $\eta_E:X\rightarrow X/E$ is linear and that this occurs iff $[0]$ is a subspace of $X$ (in which case, $E=\{(x,x')\mid x-x'\in [0]\})$.)

$\textbf{Exercise question:}$ Let $f,g:X\rightarrow Y$ be linear. Using exercise 3, (exercise in the quoted block text above) show that $\text{eq}(f,g)$ and $\text{coeq}(f,g)$ in $\textbf{Vect}.$

Can I let the set $K$ to be $K=\text{ker}(f-g),$ where $\text{ker}(f-g)=\{a\in X\mid (f-g)(a)=0\}=\{a\in X\mid f(a)=g(a)\}=\text{ker}(f).$ We define $h=eq(f-g,0):\text{ker}(f-g)\rightarrow X$ where $K\subset X$. If so,then here is my attempt at a proof.

Theorem: $\textbf{Vect}$ has equalizers.

Proof: Suppose $f,g:X\rightarrow Y$ are linear maps in the category of $\textbf{Vect},$ (category of vector space) and let $K=\text{ker}(f-g),$ where $\text{ker}(f-g)=\{a\in X\mid (f-g)(a)=0\}=\{a\in X\mid f(a)=g(a)\}.$ We define the embedding map $h:K\rightarrow X$ and since $K\subset X,$ $h(K)\subset X$ then $f\circ h =g\circ h$. Hence the map $h$ is a equalizer of the pair of maps $f,g$. Suppose there exists another map $\psi: D\rightarrow X$ in $\textbf{Vect}$ such that $f\circ \psi=g\circ \psi.$ The map $\psi$ being linear, $(f-g)(\psi)=0$, and $\psi\in K,$ along with the definition of $K$, $\psi(D)\subset K,$ hence $\text{im}(\psi)\subset K$. There exists a unique map $\psi':D\rightarrow K$ in $\textbf{Vect},$ where $\psi=h\circ \psi'.$ Indeed, if $h\circ \psi''=\psi,$ where $\psi'':D\rightarrow K$ is in $\textbf{Vect};$ then $\psi=h\circ \psi'=h\circ \psi''.$ $h$ being injective, implies that $\psi'=\psi''$, which shows uniqueness. Thus $h$ is the equalizer of the pair of maps $(f,g)$ and so $\textbf{Vect},$ has equalizers.

The above proof is adapted from "Categorical properties of locally $m-$convex algebras" by D Rosa, which appeared in: Topological vector Spaces, algebra and Related Area, by A Lau, I Tweddle.

In it, the two authors defined $L$ as the category of locally $m-$convex algebras. Here is the reproduced proof.

Theorem: $L$ has equalizers.

Proof: Given $\phi_1,\phi_2:A\rightarrow B$ in $L$, let $C=\{a\in A\mid \phi_1(a)=\phi_2(a)\}$ with the relative topology. Clearly the natural embedding $i_C:C\rightarrow A$ is in $L$ and $\phi_1\circ i_C=\phi_2\circ i_C.$ Suppose there exists $\psi:D\rightarrow A$ in $L$ such that $\phi_1\circ \psi=\phi_2\circ \psi.$ By the definition of $C,$ it follows that $\psi(D)\subset C.$ So there exists a unique $\psi':D\rightarrow C$ in $L,$ namely the corestriction of $\psi,$ such that $\psi=i_C\circ \psi'.$ Thus $i_C$ is the equalizer of the pair $(\phi_1, \phi_2),$ and so $L$ has equalizers.

I have two quick questions:

(1) Is my attempt proof correct

(2) In Lau and Tweddle's proof above they did not define what $\psi'$ suppose to be, they did justify its existence in $L$ by stating that $\psi$ factors through $i_C$ because of the corestriction of $\psi$.

However can I let $\psi'=h^{-1}\circ \psi,$ where $h^{-1}$ is the inverse image of $h,$ meaning $h^{-1}(X)=K,$ and $(h\circ h^{-1})(X)=h(h^{-1}(X))=h(K).$ Also $\psi(D)\subset X,$ then $h\circ (h^{-1}\circ \psi)=(h\circ h^{-1})\circ \psi={id}_X \circ \psi=\psi.$

Thank you in advance.

Answer 1

$(1)$ Your attempted proof is essentially correct but there are some notational errors, and some things I'd like to point out. Based on our previous interactions I think they are worth mentioning, but forgive me if it feels pedantic.

"... and since $K\subset X$, $h(K)\subset X$ then $f\circ h=g\circ h$

That sentence looks like it is trying to convey: "$f\circ h=g\circ h$ folllows from the fact $h(K)\subset X$" which is definitely not true. Rather, $f\circ h=g\circ h$ is true simply by definition of $K$: for $a\in K$, $f(h(a))=f(a)=g(a)=g(h(a))$ by definition. That is why $fh=gh$.

"The map $\psi$ being linear, $(f-g)\psi=0$"

Well, actually $(f-g)\psi$ would still be true even if $\psi$ wasn't linear.

When you say "$\psi\in K$", that doesn't make sense. Did you mean, for all $a\in D$, $\psi(a)\in K$? One last thing: you should also make sure you know why a linear map $\psi':D\to K$ exists (you got the uniqueness right) which is not something addressed in your proof.

$(2)$:

"They did not define what $\psi'$ is supposed to be"

Neither did you! But $\psi'$ is defined exactly the same by both you and the authors of that text. $\psi':D\to C$ simply takes $d\mapsto\psi(d)$: since $\phi_1(\psi(d))=\phi_2(\psi(d))$ is true, $\psi(d)$ can be viewed as an element of $C$. This is indeed "corestriction".

"However can I let $\psi'=h^{-1}\circ\psi$"

Yes and no... $h^{-1}$ doesn't necessarily exist as a function $X\to C$ and when you say "meaning $h^{-1}(X)=K$" etc. that's just not right: $h^{-1}(X)=K$ is true, but it doesn't exactly define what $h^{-1}$ is. $h^{-1}\circ\psi$ does make some sense, but only because $\psi(D)\subseteq C$, and once you've realised that, you've realised there's no need to talk about "$h^{-1}$".