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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$
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 $\psi=\psi\cc j\cc\al=0$.
