Why is $[f(a)]=[0]$ in the proof that coequalizer is onto in category $\textbf{Vect}$

Question

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

$\color{Green}{Background:}$

$\textbf{(1) Proposition:}$ In the category $\textbf{Vect}$

$\textbf{(i)}$ $f$ is an epimorphism iff $f$ is onto iff $f$ is a coequalizer.

$\textbf{(2)}$ $\textbf{Proposition:}$ Every coequalizer is an epimorphism.

Proof: If $h=\text{coeq}(p_1,p_2)$ and $k_1\circ h=k_2\circ h,$ then $$(k_1\circ k_2)\circ p_1=k_1\circ(h\circ p_2)=k_1\circ(h\circ p_2)=(k_1\circ h)\circ p_2$$ and so there is a unique morphism $\psi$ such that $k_1\circ h=\psi\circ h.$ Thus $k_1=\psi=k_2.$

$\textbf{(3)}$ $\textit{A coequalizer is an epimorphism.}$

Proof: If $f:A\to B$ $\textit{is an epimorphism, f is onto.}$ Let $E$ be the equivalence relation $\{(b,b')\mid b-b'\in f(A)\}$ on $B.$ Following a standard if sloppy convention, we write the quotient set as $B/f(A)$ rather than $B/E. B/E$ is easily checked to be a vector space with operations $[b]+[b']=[b+b'], \lambda\cdot[b]=[\lambda\cdot b].$

Define the two linear maps $$t:B\to B/f(A):b\mapsto [b]$$ $$u:B\to B/f(A):b\to [0].$$ Then $t\circ f$ and $u\circ f$ are both zero maps (since $[f(a)]=[0]$ for all $a$ in $A$) and so $t=u$ because $f$ is an epimorphism. Thus for all $b\in B, [b]=[0],b\in f(A),$ and $f$ is onto.

$\color{Red}{Questions:}$

I don't understand why in the proof, why both $[f(a)]=[0]$ and $[b]=[0].$ Is it that $[f(a)]=[0]$ is an assumed fact, if so, how and why? Hence $(b,b')\in E$ or $bEb'.$ Thank you in advance

Answer 1

Is it because of the following, if we let $b'=f(a)=0$ and $b=0,$ then $b-b'\in f(A),$ Hence $(b,b')\in E$ or $bEb'.$ Thank you in advance

No. You can't "let" $f(a)=0$ for a generic $a$ unless you know $f$ is the zero map. To show $[f(a)]=[0]$ in $B/f(A)$ (which, by the way, is not sloppy notation and is entirely unambiguous : )) - well, this should be easy. If not, I recommend you review how to work with quotient (vector) spaces. $[f(a)]=[0]$ by definition if and only if $f(a)-0\in f(A)$. So, that's if and only if $f(a)\in f(A)$, which is true. Therefore, $[f(a)]=[0]$.

After concluding that $t=u$, we know $[b]=0$ because $[b]=t(b)=u(b)=[0]$. That's if and only if $b-0\in f(A)$, i.e. $b\in f(A)$. And this is true for all $b$, so, it must be that $f$ is surjective (onto). So you conclude epimorphisms are onto.