In a dg-category $\mathcal{C}$, the $n$-translation of an object $C$ is an object $C[n]$ representing the functor $$ {\rm Hom}(-,C)[n]. $$ The cone of a closed morphism $f\colon C \to D$ of degree zero is an object ${\rm Cone}(f)$ representing the functor $$ {\rm Cofiber}\big({\rm Hom}(-,C) \stackrel{{f_{\ast}}}{\longrightarrow}{\rm Hom}(-,D)\big). $$ On the other hand, the homotopy cofiber of $f$ is an object ${\rm Cofiber}(f)$ representing the functor $$ {\rm Fiber}\big({\rm Hom}(D,-) \stackrel{{f^{\ast}}}{\longrightarrow}{\rm Hom}(C,-)\big). $$

Now, suppose $\mathcal{C}$ has zero object, all translations of all objects, and all cones of all morphisms.

My question: Is there any easy way to show ${\rm Cofiber}(f)$ and ${\rm Cone}(f)$ are isomorphic?

To be more efficiently, I know that there are canonical closed morphism $\iota\colon D\to{\rm Cone}(f)$ of degree $0$ and morphism $h\colon C\to{\rm Cone}{f}$ of degree $-1$. They induce a natural cochain map $$ {\rm Hom}\big({\rm Cone}(f),-\big) \longrightarrow {\rm Fiber}\big({\rm Hom}(D,-) \stackrel{{f^{\ast}}}{\longrightarrow}{\rm Hom}(C,-)\big) $$ which sends any $x\colon{\rm Cone}(f)\to X$ to the pair $(x\circ\iota,x\circ h)$.

However, I don't know how to finish the proof, i.e. show this is an isomorphism.


1 Answer 1


A very short answer would be as follows. What you define to be $\mathrm{Cone}(f)$ lies in a triangle in $\mathcal C$: \begin{equation} C \xrightarrow{f} D \to \mathrm{Cone}(f), \end{equation} whereas what you define as $\mathrm{Cofiber}(f)$ lies in a triangle in $\mathcal C^{\mathrm{op}}$: \begin{equation} \mathrm{Cofiber}(f)[-1] \to D \xrightarrow{f^\mathrm{op}} C. \end{equation} This triangle in $\mathcal C^\mathrm{op}$ is the same as the previous triangle in $\mathcal C$.


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