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Kernel pairs can be taken in any category with pullbacks, when there is a zero object we also have kernels. Then there is a morphism from the kernel to the kernel pair (via pullback uniqueness).

What restrictions do we need to put on the category to make this an isomorphism?

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up vote 4 down vote accepted

Let's call our map $f:X \to Y$, the kernel $K$, and the kernel pair $P$. As you said, the natural map from the kernel to the kernel pair comes from a map of pullback squares. If this has an inverse, then chasing this diagram shows that the map $P \to Y$ factors through zero. But the map $X \to Y$ factors through $P$, since $P$ is the pullback of $X \to Y \leftarrow X$ and there is also a commutative square

X -> X
|    |
v    v
X -> Y

where the top and left-hand maps are the identity.

Thus the map $X \to Y$ is zero. Your condition only holds in categories equivalent to the terminal category!

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I can see that if $X\rightarrow Y$ is the pullback of $X\rightarrow Y\leftarrow X$ then it must be the zero morphism, and hence your conclusion. But I don't see why it is the pullback - there's no morphism from $Y\rightarrow X$? – Mozibur Ullah Feb 8 '13 at 7:48
Sorry, my phrasing was ambiguous. I meant that the kernel pair $P$ is the pullback of $X \to Y \leftarrow X$, by definition, and thus $X \to Y$ factors through $P$. I'll edit my answer. – Paul VanKoughnett Feb 12 '13 at 6:14

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