nForum - Search Results Feed (Tag: relative) 2022-10-04T07:12:31-04:00 https://nforum.ncatlab.org/ Lussumo Vanilla & Feed Publisher relative adjoint https://nforum.ncatlab.org/discussion/3515/ 2012-01-31T12:49:55-05:00 2016-09-18T03:31:48-04:00 eparejatobes https://nforum.ncatlab.org/account/373/ Hi everyone! I’ve created relative adjoint functor, and linked to it from the local definition of adjoint functor (a partially defined adjoint yields an adjoint relative to the inclusion of a full ...

Hi everyone!

I’ve created relative adjoint functor, and linked to it from the local definition of adjoint functor (a partially defined adjoint yields an adjoint relative to the inclusion of a full subcategory).

$L {\,\,}_J\!\dashv R$ and $L \dashv_J R$ is as far as I know nonstandard notation, but I think it’s ok, even if the left subscript feels a bit kludgy. I will add more stuff in the next few days.

PS: Thanks a lot to all the nLab contributors; in the past few years I’ve learn a lot through here :) I now have the time and a little bit of confidence to contribute, so any pointers, tips, formatting, style suggestions, whatever will be greatly appreciated

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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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