# Difficult category theory: kernels

In a category with a zero object ($0$ = initial and terminal) and zero morphisms are unique $A \to 0 \to B$ for every $A,B$, define the kernel of a map as the equalizer of $(f,0)$ and cokernel dually.

I try to prove that if $f = \ker(g)$ then $f = \ker(\operatorname{coker}(f))$. So set $c = \operatorname{coker}(f)$. We have the diagrams

$$A \xrightarrow{f} B \xrightarrow{g} C$$ $$A' \xrightarrow{\ker c} B \xrightarrow{c} C'$$

1. $f$ equalizes $(g,0)$: $g f = 0$
2. $c$ coequalizes $(f,0)$: $c f = 0$
3. $\ker c$ equalizes $(c,0)$: $c \ker(c) = 0$

By (2) and $0 = 0 f$ we get that $f$ claims to equalize $(c,0)$ so there exists a universal map $u : A \to A'$ such that $\ker(c) u = f$, therefore $f$ equalizes $(c,0)$ and by uniqueness of equalizers $f = \ker(c)$.

But I don't know if this is right, am I making some mistakes and can anyone show me a simple proof instead? Thank you

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The kernel of a morphism should be an object, not a morphism. –  Tobias Kildetoft Jan 16 '13 at 19:14
@Tobias: In category theory, a kernel is a morphism. –  Clive Newstead Jan 16 '13 at 19:16
@CliveNewstead Woops, my bad. But it should be not just a morphism, but an object with an associated morphism (the terminal object such that the original morphism factors through this new object with this new morphism). –  Tobias Kildetoft Jan 16 '13 at 19:19
@Tobias Technically, while it is possible in a category that $\operatorname{Hom}(A,B)$ can have common elements with $\operatorname{Home}(X,Y)$, most category theory treats them as disjoint - that is, morphisms come with a distinct "from" and "to" object. –  Thomas Andrews Jan 16 '13 at 19:34

What you've written is mostly correct, but I'm not very convinced by your 'uniqueness of equalizers' argument. In any case, it could be expressed more neatly. What I'd do is as follows.

Let $f : A \to B$ be the kernel of $g : B \to C$. Let $c = \operatorname{coker}(f) : B \to Q$. We want to show that $f=\ker(c)$, which amounts to showing that $f$ equalizes $c$ and $0$. So suppose $p : P \to B$ has $cp=0$.

Since $cf=gf=0$ and $c$ is a coequalizer, there is a unique $q : Q \to C$ with $qc=g$. But then $0=q0=qcp=gp$, so $gp=gf=0$ and so there is a unique $u : P \to A$ with $p=fu$. Thus $f$ equalizes $c$ and $0$, so $f = \ker(c)$, as desired.

This is illustrated in the following makeshift commutative diagram.

$$\begin{array}{ccccccccc} &&P&&&&&& \\ & \overset{u}{\swarrow} & & \overset{p}{\searrow} &&&&& \\ A && \xrightarrow{f} && B && \xrightarrow{g} && C \\ &&&&& \underset{c}{\searrow} && \underset{q}{\nearrow} & \\ &&&&&& Q & \end{array}$$

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Thank you. What do you mean by $cf = gf = 0$? since they have different codomain –  user58512 Jan 16 '13 at 19:53
@user58512: Good spot. I abused notation; it means that $cf=c0$ and $gf=g0$. Likewise with $gp=gf=0$, what I 'meant' was $gp=0p$ and $gf=0f$. –  Clive Newstead Jan 16 '13 at 19:55
thank you ! aaa –  user58512 Jan 16 '13 at 19:58