Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 3 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!

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.