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.