Mapping cone is unique up to (non-unique) isomorphism

Question

I am reading about triangulated categories. According to the axioms, given a morphism $f : A \to B$, there is a so-called mapping cone $C$ of $f$, i.e., an object $C$ with maps $g: B \to C$ and $h: C \to A[1]$ such that $A \to B \to C \to A[1]$ is a distinguished triangle. $C$ is not unique, but on Wikipedia it is claimed that any two mapping cones $C$ and $C'$ for $f$ are isomorphic. Now, I can see that by the axioms there must be morphisms $j:C \to C'$ and $k:C' \to C$ such that $(k \circ j) \circ g = g$ and other similar equalities, but I can really not deduce that $k \circ j=\mathrm{id}_C$. Any hints?

Answer 1

The point is that $Hom(X,.)$ sends a distinguished triangle to a long exact sequence.

Now consider a morphism of triangles $$\require{AMScd} \begin{CD} A@>>> B@>>> C@>>>A[1]\\ @VVV@VVV@VVV@VVV\\ A'@>>>B'@>>> C'@>>> A'[1] \end{CD} $$ such that two out of the three maps are isomorphisms, say $A\to A'$ and $B\to B'$. We want to show that $C\to C'$ is also an isomorphism.

For any $X$, apply the functor $Hom(X,.)$. You get two long exact sequences and a morphism between them such $Hom(X,A[n])\to Hom(X,A'[n])$ and $Hom(X,B[n])\to Hom(X,B'[n])$ are isomorphisms. From the 5-lemma you have that for any $X$, the map $Hom(X,C)\to Hom(X, C')$ is an isomorphism. By naturality, you can deduce that $Hom(.,C)\to Hom(.,C')$ is a natural isomorphism. Then from Yoneda's lemma, $C\to C'$ is an isomorphism.

The unicity of mapping cone follows from this.