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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.

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What definition of abelian category are you using? –  Zhen Lin Nov 1 '13 at 18:54
    
An abelian category is an additive category such that every morphism has kernel and cokernel, every monic is kernel of its cokernel, every epic is cokernel of its kernel. This is the definition given by Grothendieck in Tohoku. –  A.G Nov 1 '13 at 19:29

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up vote 3 down vote accepted

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.

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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).

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