# Why is in the category of pointed sets not every epimorphism a cokernel?

A question in Tennison's Sheaf Theory is about the category of pointed sets and its characteristics. I have that

• its zero object is given by $(\{x\},x)$
• the kernel of $f\colon (A,a)\to (B,b)$ is given by $(f^{-1}(b),a)$
• the cokernel is given by $(f(A),b)$
• epimorphisms are surjective maps

but I fail to see why this breaks down cokernels.

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What do you mean by "this breaks down cokernels"? What does "this" refer to, and what's breaking down? –  Alon Amit Jun 17 '11 at 9:26
Am I wrong? It seems that the cokernel is $(B-f(A)\cup\{b\},b)$. –  wxu Jun 17 '11 at 9:35
It sounds like surjectivity breaks for cokernels. No, a cokernel is a colimit, then an epimorphism, then it is surjective. –  beroal Jun 17 '11 at 18:26
@Alon: It means not every epimorphism (giving a quotient object) is a cokernel. So the "this" refers to "what the (co)kernels and epis of the category of pointed sets are". But it might've been a colourful and informal description :). –  Pieter Jun 19 '11 at 11:41
@wxu: You're right, and that was why I couldn't see the obvious. As @beroal mentioned, the surjectivity fails in the case of split epimorphisms. Thanks for the answers! –  Pieter Jun 19 '11 at 11:42