# How to show mapping cones are homotopy cofibers

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.

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