Does $X\overset{f}{\rightarrow} Y\rightarrow0$ being exact imply $\operatorname{coker}f=0$?

In a category with zero object, it is easy to see that if $0\rightarrow X\overset{f}{\rightarrow} Y$ is exact then $\ker f=0$, since $0\rightarrow X$ is monic and hence is its own image. However, when I try to prove its dual statement, that if $X\overset{f}{\rightarrow}Y\rightarrow0$ is exact then $\operatorname{coker} f=0$, somehow I can't. Am I overlooking certain fact, or is it true at all? Or is it a corroboration of my mounting suspicion that things are not exactly symmetrical between a category and its dual category?

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You've already proved the cokernel thing too! It does follow by duality from the statement you proved for the kernel. It might help to dualize your proof step by step to see what's going on. – Omar Antolín-Camarena Nov 30 '12 at 14:25
The exactness of $X\overset{f}{\rightarrow} Y\rightarrow0$ implies that $\operatorname{id}_Y:Y\rightarrow Y$ is the image of $f$, so provided that the category has epimorphic images, then it follows that $f$ is an epimorphism. But I can't seem to go anywhere without the assumption. – ashpool Nov 30 '12 at 14:44

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Proposition $\$ Let $$X\rf{\phi}Y\ra0$$ be an exact sequence in a category with kernels. Then $\cok\phi=0$.
Proof We will show that $Y\ra0$ is a cokernel of $\phi$. This is equivalent to showing that if $\psi:Y\ra Z$ is a morphism such that $\psi\cc\phi=0$, then $\psi=0$. So let $\psi:Y\ra Z$ be such a morphism, and $j:\krn\psi\ra Y$ its kernel. Since $\psi\cc\phi=0$, there exists a morphism $\hphi:X\dra\krn\psi$ such that $j\cc\hphi=\phi$. $$\xymatrix{& \krn\psi \ar@{>->}[d]^j \\ \ar@{-->}[ru]^{\hphi} X \ar[r]^\phi & Y \ar[r] \ar[d]_\psi & 0 \ar@{-->}[ld] \\ & Z}$$ By exactness, $\id_Y:Y\ra Y$ is an image of $\phi$. Hence, there exists a morphism $\al:Y\dra\krn\psi$ such that $j\cc\al=\id_Y$. Then $\psi=\psi\cc j\cc\al=0$.