nForum - Search Results Feed (Tag: abelian) 2021-12-03T05:58:53-05:00 https://nforum.ncatlab.org/ Lussumo Vanilla & Feed Publisher Kernel of split epimorphism is cokernel of right-inverse? https://nforum.ncatlab.org/discussion/4965/ 2013-06-01T01:55:13-04:00 2013-06-13T06:25:32-04:00 joe.hannon https://nforum.ncatlab.org/account/887/ In reduced homology#relation to relative homology, at the bottom is a computation where one step is that kerH 0(&epsi;)&cong;cokerH 0(x).\ker H_0(\epsilon) \cong \operatorname{coker}{H_0(x)}. ...

In reduced homology#relation to relative homology, at the bottom is a computation where one step is that $\ker H_0(\epsilon) \cong \operatorname{coker}{H_0(x)}.$ Can you explain that step to me? If you think of kernels and cokernels as morphisms, then there’s no way such an isomorphism can hold, since $\ker H_0(\epsilon)$ is a morphism into $H_0(X)$, while $\operatorname{coker}{H_0(x)}$ is a morphism out of $H_0(X).$ But if you think of kernels and cokernels as objects, then I guess it’s possible for them to be isomorphic. My guess is it must follow from the fact that $H_0(\epsilon)\circ H_0(x)$ is an iso, so the statement is something like “the kernel of a split epimorphism is the cokernel of its right-inverse,” but I can’t figure it out.

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