User:Jo be

Proposition

Let $\mathcal{C}$ be an abelian category, and let $f\in\mathcal{C}\left[X, Y\right]$ be a morphism. Let $\mathrm{ker}\,f\in\mathcal{C}\left[K,X\right]$ be the kernel of $f$, that is, the equalizer of $f$ and the zero morphism $0_{X,Y}$.

Then $f$ is a monomorphism if and only if $K$ is the zero object

Proof

We first prove that monomorphism implies trivial kernel.

Suppose there is some morphism $k^\prime\in\mathcal{C}\left[K^\prime, X\right]$ that makes $f$ and $0_{X,Y}$ equal. Since $\mathcal{C}$ is abelian, all kernels exist, and thus by the definition of kernel as equalizer, there $\exists ! m\in\mathcal{C}\left[K^\prime, K\right]$ such that $k^\prime = \ker f\circ m$.

Note that $f\circ k^\prime = 0_{X,Y}\circ k^\prime = 0_{K^\prime,Y} = f\circ 0_{K^\prime,X}$. By assumption $f$ is a monomorphism, so this implies $k^\prime = 0_{K^\prime,X}$. But every zero morphism factors uniquely through the zero object, so $K=0$.

Next, we show that $K=0$ implies that $f$ is monic.

Suppose $f\circ g_1 = f\circ g_2$ for some $g_i\in\mathcal{C}\left[A,X\right], i=1,2$. Then, $f\circ(g_1 - g_2) = f\circ g_1 - f\circ g_2 = 0_{A,Y}=0_{X,Y} \circ (g_1-g_2)\,\therefore g_1-g_2$ makes $f$ and the zero morphism equal. $K=0 \therefore \exists! m:\,g_1-g_2=0_{0,X}\circ m=0_{0,X}\circ 0_{A,0}=0_{A,X}$, so $g_1-g_2=0_{A,X}\iff g_1=g_2$, since zero morphisms are unique.

$\Box$

Proof of Uniqueness of Representing Objects

By the Yoneda Embedding Theorem, the functor $h_- \colon C \to [C^{op}, \mathbf{Set}]$, $X \mapsto \operatorname{Hom}(X, -)$ is fully faithful. Now, by assumption, we have a natural isomorphism $\zeta \colon \operatorname{Hom}(A, -) \xrightarrow{\eta} F \xrightarrow{\xi^{-1}} \operatorname{Hom}(B, -)$. Since fully faithful functors reflect isomorphisms, we obtain an isomorphism $f\colon A \to B$ in $C$, which moreover is compatible with $\zeta$ in the sense that for any object $X$ in $C$ and any morphism $g \in \operatorname{Hom}(B, X)$, the equality $\zeta^{-1}(g) = g \circ f = h^f(g)$ holds. Thus, $h^f = \eta^{-1} \circ \xi$, or equivalently, $\xi = \eta \circ h^f$.