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. Using the following exercise (one in the block quote), show that $\text{coeq}(f,g)$ exists, where $\text{coeq}(f,g)$ denotes the coequalizer of $(f,g)$. I also posted the same question, but for showing that equalizer in $\textbf{Vect}$ exists.
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 $\textbf{Vect}.$
Let the vector subspace $F=\text{Im}((f-g)(X))$ be defined as $\text{Im}((f-g)(X))=\{f(x)-g(x)\mid x\in X\},$ where $f,g:X\rightarrow Y.$ We also define $h=\text{coeq}(f-g,0):Y\rightarrow Y/F=Y/{\text{coeq}(f,g)}.$ If so,then here is my attempt at a proof.
Theorem: $\textbf{Vect}$ has coequalizers
Proof: Suppose $f,g:X\rightarrow Y$ are linear maps in the category of $\textbf{Vect}$. Let $F$ be the subspace $F=\text{Im}((f-g)(X))=\{f(x)-g(x)\mid x\in X\},$ and let $E$ be the equivalence relation generated by $F$ where $E=\{(y,y')\mid [y]_F=[y']_F\in Y/F\}=\{(y,y')\mid [y]_F-[y']_F=[0]_F\in Y/F\}.$ Then the canonical quotient map $w:Y\rightarrow Y/F$ defined as $w(y)=[y]_F,$ is in $\textbf{Vect}$ and $w\circ f=w\circ g.$ Suppose there exists another linear map $\psi:Y\rightarrow D$ defined as $\psi(y)=d,$ in $\textbf{Vect}$ such that $\psi\circ f=\psi\circ g.$ Since $\psi$ is linear, and $w\circ f-w\circ g=w\circ(f-g)=0,$ then $(f-g)\in \text{ker}(w),$ $F\subset \text{ker}(w),$ and $E\subset \text{ker}(w).$ We assume that $\psi$ is compatible with $F,$ meaning for all $x, x'\in Y,$ $xFx'\Longrightarrow \psi(x)=\psi'(x').$ So let $A'=[y]_F=Y/F,$ and let $w'=\{(A',d)\in Y/F\times D\mid \exists y\in A' (\psi(y)=d)\}$ We need to show the existence of $w':Y/F\rightarrow D$ in $\textbf{Vect}.$ So suppose $A'\in Y/F.$ Let $y\in A'$ such that $A'=[y]_F.$ Let $\psi(y)=d.$ Since $y\in [y]_F=A'$ and $\psi(y)=d,$ we have $(A',d)\in w'.$ For showing $d$ being unique, suppose $(A',d')\in w'.$ Then we can let $y'\in A'$ such that $\psi(y')=d'.$ Since $y'\in A'=[y]_F,$ $yFy',$ and since $\psi$ is compatible with $F,$ $d'=\psi(y)=\psi(y')=d,$ which implies $w'(w(y))=w'([y]_F)=\psi(y).$
To show that $w'$ is unique, suppose that $w'':Y/F\to D$ and for all $y\in A', w''(w(y))=w''([y]_F)=\psi(y).$ Suppose $A'\in Y/F,$ then we can pick some $y\in A'$ where $A'=[y]_F.$ Therefore $w''(A')=w''(w(y))=w''([y]_F)=\psi(y)=w'([y]_F)=w'(w(y))=w'(A').$ Since $A'$ was arbitrary, we have $w''=w',$ and we have shown that $w'(w(y))=w'([y]_F)=\psi(y).$ Then $w'\circ w=\psi.$ Thus $w$ is the coequalizer of the pair of linear maps $(f,g)$ and so $\textbf{Vect}$ has coequalizers.
The above proof is adapted from how coequalizer are determined in the case of abelian groups, and also from a theorem in an article titled: "Categorical properties of locally $m-$convex algebras" by D Rosa, which appeared in the text: Topological vector Spaces, algebra and Related Area, by A Lau, I Tweddle.
In that article, the two authors defined $L$ as the category of locally $m-$convex algebras. Here is the reproduced proof.
Theorem: $L$ has coequalizers.
Proof: Given $\psi_1, \psi_2:A\rightarrow B$ in $\textbf{L},$ let $I$ be the closure of the ideal generated by $\{\psi_1(a)-\psi_2(a):a\in A\}.$ Then the canonical quotient map $Q:B\rightarrow B/I$ is in $\textbf{L}$ and $Q\circ \psi_1=Q\circ \psi_2.$ Suppose there exists $\psi:B\rightarrow D$ in $\textbf{L},$ such that $\psi\circ \phi_1=\psi\circ \phi_2.$ Since $\psi$ is continuous it follows that $I\subset \text{ker}(\psi).$ Hence there exists a unique $\psi':B/I\rightarrow D$ in $\textbf{L},$ such that $\psi'\circ Q=\psi.$ Thus $Q$ is the coequalizer of the pair $(f,g)$ and in $\textbf{L},$ has coequalizers.
Also for constructing the coequalizer in the category of $\textbf{Grp}$ is a solution to the following question (from Harold Simmons' category theory text):
Consider a parallel pair of morphisms $f,g:A\to B$ between groups (written multiplicatively) Let $$F=\{f(a)g(a)^{-1}\mid a\in A\}$$
and let $K$ be the normal subgroup generated by $F.$ Show that the canonical quotient $B\to B/K$ is the coequalizer of $f$ and $g$ in $\textbf{Grp}$
Proof: Given $f,g:A\to B\xrightarrow{k}B/K$
where $K$ is the canonical quotient. For each $a\in A$ we have
$$k(f(a)g(a)^{-1})=k(1)=1$$
which leads to $$k(f(a))=k(g(a))$$
and hence $k$ does make equal $f$ and $g$. We show that $k$ is the coequalizer of $f$ and $g$.
Consider the group morphism $$B\xrightarrow{h}X$$
which does make equal $f$ and $g$. For each $a\in A$ we have $$h(f(a))=h(g(a))$$
so that $$h(f(a)g(a)^{-1})=h(1)=1$$
and hence $$f(a)g(a)^{-1}\in {ker}(h)$$
to show $$F\subset {ker}(h)$$
and hence $$K\subset {ker}(h)$$
by the construction of $K.$ There is a unique morphism $m$ for which $h=m\circ k$
I have quick question:
Is my proof correct? In particular, I used the idea of ideal of $I$ being the equivalence relation generated by the direct image of $f-g$. I hope it matches with the equivalence relation $E$ in the quoted exercise. Also, I know $Y/F$ has to do with the cokernel of $w$, but I did not mentioned that.
Thank you in advance.
Edit: I found out there was some typos and made the necessary corrections, and also, I was able to figure out how to adapt or imitate the idea/technique from the case of abelian groups.
Like with the other post the proof is essentially right but has some confusions and gaps.
As noted by another user in another post of yours, the often-quoted "exercise $3$" is wrong ("has a linear structure iff. $[0]$ is a subspace..."). All that you need to care about is that, given any subspace $K$ of $U$, there is a quotient $U/K$.
This is mostly true, but has absolutely nothing to do with $\psi$. $F=\ker w$ exactly, not just $F\subset\ker w$, and $E\subset\ker w$ is wrong. $E$ is a subset of $Y\times Y$ and $\ker w$ is a subset of $Y$.
Then your "hence there exists a unique $w'$..." is coming from nowhere. You have not actually made any deductions from the fact $\psi f=\psi g$.
By copying the proof from the other text you seem to have lost sight of what's going on. I strongly recommend thinking through each sentence of the other text for yourself so you know exactly why it is true. Maybe remind yourself of the "universal property of the quotient map" for set-based categories: that's basically what you're trying to prove here. You know $\psi f=\psi g$, now you need to tell the reader exactly why that implies there is a (unique) linear map $\psi':Y/F\to D$ satisfying $\psi'\circ w=\psi$. You haven't done that yet (and that is arguably the main point of the proof).
Some thoughts for you:
How should we define $\psi'([y])$ for some equivalence class $[y]\in Y/F$? You'd like to set it equal to $\psi'([y]):=\psi(y)$ (why do we want that?) but there's a problem with well-definedness: what if $[y]=[y']$, can we then be sure that $\psi(y')=:\psi'([y'])=\psi'([y])=\psi(y)$?