[proofplan] The assertion is exactly the normality and conormality axiom in the definition of an abelian category. We first apply the axiom to a monomorphism and its chosen cokernel, then use the universal property of kernels to identify any two resulting kernel morphisms by a unique isomorphism. The epimorphism statement is the dual argument, using cokernels and their universal property. [/proofplan] [step:Apply the abelian category axiom to the monomorphism] Let $m: A \to B$ be a monomorphism in $\mathcal{A}$, and let $q: B \to Q$ be a cokernel of $m$. Since $\mathcal{A}$ is abelian, every monomorphism is normal: that is, every monomorphism is a kernel of its cokernel. Therefore $m$ is a kernel of $q$. Equivalently, $q \circ m = 0$, and for every object $X$ of $\mathcal{A}$ and every morphism $f: X \to B$ satisfying $q \circ f = 0$, there exists a unique morphism $\tilde{f}: X \to A$ such that \begin{align*} m \circ \tilde{f} = f. \end{align*} Thus $m$ represents the subobject of $B$ annihilated by $q$. [/step] [step:Identify any chosen kernel of the cokernel with the original monomorphism] Let $i: K \to B$ be any chosen kernel of $q$. Since both $m: A \to B$ and $i: K \to B$ are kernels of the same morphism $q: B \to Q$, their universal properties give unique comparison morphisms \begin{align*} u &: A \to K, \\ w &: K \to A \end{align*} such that \begin{align*} i \circ u &= m, \\ m \circ w &= i. \end{align*} We show that $u$ and $w$ are inverse isomorphisms. Composing the first identity with $w$ gives \begin{align*} m \circ w \circ u = i \circ u = m. \end{align*} Also \begin{align*} m \circ \operatorname{id}_A = m. \end{align*} Since $m$ is a kernel of $q$, the factorisation of $m: A \to B$ through $m: A \to B$ is unique. Hence \begin{align*} w \circ u = \operatorname{id}_A. \end{align*} Similarly, composing the second identity with $u$ gives \begin{align*} i \circ u \circ w = m \circ w = i, \end{align*} while \begin{align*} i \circ \operatorname{id}_K = i. \end{align*} Since $i$ is a kernel of $q$, the factorisation of $i: K \to B$ through $i: K \to B$ is unique. Hence \begin{align*} u \circ w = \operatorname{id}_K. \end{align*} Therefore $u: A \to K$ is an isomorphism, and the condition $i \circ u = m$ determines it uniquely by the universal property of $i$. [/step] [step:Apply the dual abelian category axiom to the epimorphism] Let $e: A \to B$ be an epimorphism in $\mathcal{A}$, and let $k: K \to A$ be a kernel of $e$. Since $\mathcal{A}$ is abelian, every epimorphism is conormal: that is, every epimorphism is a cokernel of its kernel. Therefore $e$ is a cokernel of $k$. Equivalently, $e \circ k = 0$, and for every object $Y$ of $\mathcal{A}$ and every morphism $g: A \to Y$ satisfying $g \circ k = 0$, there exists a unique morphism $\tilde{g}: B \to Y$ such that \begin{align*} \tilde{g} \circ e = g. \end{align*} Thus $e$ represents the quotient of $A$ by the subobject represented by $k$. [/step] [step:Identify any chosen cokernel of the kernel with the original epimorphism] Let $p: A \to C$ be any chosen cokernel of $k$. Since both $e: A \to B$ and $p: A \to C$ are cokernels of the same morphism $k: K \to A$, their universal properties give unique comparison morphisms \begin{align*} v &: C \to B, \\ r &: B \to C \end{align*} such that \begin{align*} v \circ p &= e, \\ r \circ e &= p. \end{align*} We show that $v$ and $r$ are inverse isomorphisms. Composing the first identity with $r$ gives \begin{align*} r \circ v \circ p = r \circ e = p. \end{align*} Also \begin{align*} \operatorname{id}_C \circ p = p. \end{align*} Since $p$ is a cokernel of $k$, the factorisation of $p: A \to C$ through $p: A \to C$ is unique. Hence \begin{align*} r \circ v = \operatorname{id}_C. \end{align*} Similarly, composing the second identity with $v$ gives \begin{align*} v \circ r \circ e = v \circ p = e, \end{align*} while \begin{align*} \operatorname{id}_B \circ e = e. \end{align*} Since $e$ is a cokernel of $k$, the factorisation of $e: A \to B$ through $e: A \to B$ is unique. Hence \begin{align*} v \circ r = \operatorname{id}_B. \end{align*} Therefore $v: C \to B$ is an isomorphism, and the condition $v \circ p = e$ determines it uniquely by the universal property of $p$. This proves both assertions. [/step]