Some questions on abelian category Let  $f: C \longrightarrow D$ be a morphism in an abelian category $\mathfrak{A}$ with kernel and cokernel both zero. How can I show that it is an isomorphism? I am not able to find it's inverse. 
Let $g: C \longrightarrow D$ be a morphism in the abelian category $\mathfrak{A}$. Let $i: K \hookrightarrow C$ be the kernel of $g$. How can I prove that the induced map $\text{coker}\ i \rightarrow D$ is a monic? 
I don't want to use Mitchell's embedding theorem.
 A: Suppose $f : C \to D$ is a morphism with zero kernel. Then $f$ must be monic: for, given any $c_0, c_1$ such that $f \circ c_0 = f \circ c_1$, we have $f \circ (c_0 - c_1) = 0$; but $\ker f = 0$, so $c_0 - c_1 = 0$, so $c_0 = c_1$. Dually, if a morphism has zero cokernel, then it must be epic. $\DeclareMathOperator{\coker}{coker}$
Next, suppose $f : C \to D$ is monic and epic. Then $f = \coker \ker f$. But then $f \circ \ker f = 0$, so $\ker f = 0$. Hence $f : C \to D$ is an isomorphism (because $\coker 0$ is always an isomorphism).
Finally, let $f : C \to D$ be any morphism. Let $\ker f : K \to C$ be the kernel, let $\coker \ker f : C \to I$ be its cokernel, and let $g : I \to D$ be the unique morphism such that $g \circ \coker \ker f = f$. Suppose $g \circ x = 0$ for some $x : X \to I$. Consider $\coker x : I \to Y$. Then there is a unique $y : Y \to I$ such that $y \circ \coker x = g$. The composite $\coker x \circ \coker \ker f : C \to Y$ is an epimorphism, hence
$$\coker x \circ \coker \ker f = \coker h$$
for some $h : Z \to C$. Thus,
$$f \circ h = g \circ \coker \ker f \circ h = y \circ \coker x \circ \coker \ker f \circ h = r \circ \coker h \circ h = 0$$
and therefore there is a unique $l : Z \to K$ such that $\ker f \circ l = h$. Now,
$$\coker \ker f \circ h = \coker \ker f \circ \ker f \circ l = 0$$
so there is a unique $m : Y \to C$ such that $m \circ \coker h = \coker \ker f$. But
$$m \circ \coker h = m \circ \coker x \circ \coker \ker f = \coker \ker f$$
and $\coker \ker f$ is epic, so $m \circ \coker x = \mathrm{id}_I$. But then $\coker x : I \to Y$ is (split) monic, hence, is an isomorphism. But that implies $x = 0$. Hence, $\ker g = 0$, and therefore $g : I \to D$ is indeed monic.
A: Let $f$ be a morphism with kernel and cokernel $0$, i.e. $f$ is monic and epic, hence a monic cokernel. Now observe that in an arbitrary linear category, any monic cokernel is an isomorphism (if $p$ is the cokernel of $i$, and monic, then $pi=0$ shows that $i=0$, and the cokernel of $0$ is an isomorphism).
