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I was wondering is there an isomorphism $$\mathrm{Hom}\big(\mathrm{cone}(A\xrightarrow{f}B),\,C\big)\,\cong\,\mathrm{cone}\big(\mathrm{Hom}(B,\,C)\xrightarrow{f^\ast}\mathrm{Hom}(A,\,C)\big)\,?$$ On the level of graded vector spaces, one has the following degree one isomorphism: $$\mathrm{Hom}(A[1]\oplus B,\,C)\,\cong\,\mathrm{Hom}(A[1],\,C)\oplus \mathrm{Hom}(B,\,C)\,\cong\,\mathrm{Hom}(A,\,C)\oplus \mathrm{Hom}(B,\,C)[1]\,.$$ I tried to check the differentials, but the signs didn't match.

If the answer of the question above is positive, are there any conceptual reasons behind that? I guess mapping cone can be viewed as "homotopical cokernel".

Any helps or comments would be very appreciated.

Yining Zhang
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  • There is a shift you seemed to have noticed but forgot : $Hom(A[1],C)=Hom(A,C)[-1]$. You should instead have $Hom(cone(f),C)=cone(f^*)[-1]$. The "conceptual reason" here is that cones are "homotopical cokernal" as you said and $cone[-1]$ is instead the homotopical kernel. – Roland Jun 06 '20 at 10:54
  • Thanks. What I meant is an isomorphism of degree +1. I guess shifting would "rotate" the exact triangle, then transfer the cokernel to kernel. Is it what you said? – Yining Zhang Jun 06 '20 at 22:10
  • I said that you missed the $[-1]$ because you said the signs didn't match and it may have come from this since the $[-1]$ put signs in the differentials. And about the conceptual reason, in (non-homotopical) category category, you have $Hom(\operatorname{coker}(f),X)=\ker(f^*)$. I said that the same is true in this homotopical context when you replace the cokernel by the cone and the kernel by $cone[-1]$ which is sometimes called the cocone. – Roland Jun 07 '20 at 07:53

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