Edit: Version 2:


Suppose that $A,B,C$ are chain complexes and $f: A \rightarrow B$ is a chain map. Suppose that there is a homotopy equivalence


$$ \text{Cone}(f: A \rightarrow B) \simeq C.$$


The chain map $u: \text{Cone}(f) \rightarrow C$ provided by the homotopy in particular means that I am provided with a chain map $g: B \rightarrow C$.


Consider now $\text{Cone}(g: B \rightarrow C)$. I would like to say that there is a homotopy equivalence between $A[-1]$ and $\text{Cone}(g)$.


Similarly, the map in the other direction $v: C\rightarrow \text{Cone}(f)$ provides me with a chain map $h: C[1] \rightarrow A$ and I would like to say that $\text{Cone}(h)$ and $B$ are homotopy equivalent.


Are these statements true and if so how can one prove them?



Homotopy equivalences and Mapping Cones
It is true. It is not trivial (this is an opinion), however it is standard. In any triangulated category, two objects and a map determine the third, up to (usually non unique) isomorphism. And the category of Chain complexes with maps up to homotopy is a classical example of triangulated category. A reference for those properties could be Happel's book (triang cat in rep. of finite dim alg).


The relevant fact is the following: (Axiom 2) if $(X\overset{u}{\to}Y\overset{v}{\to}Z\overset{w}{\to}X[-1])$ is a triangle (for chain complexes, I use the convention that $d(C\_n)\subseteq C\_{n-1}$), then $(Y\overset{v}{\to}Z\overset{w}{\to}X[-1]
\overset{-u}{\longrightarrow}Y[-1]) $ and $(Z[1]\overset{-w}{\longrightarrow}
X\overset{u}{\longrightarrow}
Y\overset{v}{\longrightarrow}Z)$ are triangles too. (Hence, if and only if.)


You should do the exercise that in the category of chain complexes with maps up to homotopy, the class of triangles being the u-ples that are isomorphic (in the homotopy category) to ones of the form $(X\overset{u}{\to}Y\overset{i}{\to}Co(u)\overset{p}{\to}X[-1])$ actually satisfies Axiom 2. This exercise is very instructive because you will find, doing it, that it is not possible in general to have all commutative squares in the category of complexes. When trying to compare the ''rotated'' triangle with a standard one, one is forced to choose maps, and there are two squares to look at: if one of them is commutative, the other is not... but there is an ''obvious'' homotopy and the square commute up to homotopy. In this way, you will learn why the category of chain complexes and chain map is not triangulated, and why the ''up to homotopy'' helps.


After that, you should do the exercise that in a triangulated category, a map $u:X\to Y$ is an isomorphism if and only if $X\overset{u}{\to} Y\to 0\to X[-1]$ is a triangle. Next, you do as exercise that if
$A\overset{f}{\to} B\overset{g}{\to} C\to A[-1]$ and
$A\overset{f}{\to} B\overset{h}{\to} D\to A[-1]$ are two triangles in a triangulated category, then $C\cong D$ in that category. Actually, you can do this in any pre-triangulated category, that is, you don't need to use the octahedral axiom. Once you are happy with this result, you should do the exercise that the category of complexes and maps up to homotopy satifies the rest of the axioms of a triangulated category (octahedral axiom is optional for this question). Also notice that proving the validity of the axioms is not the same as proving directly what you asked at the first time, to check the validity of the axioms is a kind of easy and direct.


I think this should answer your question, or if not, should give you the way that your answer should be re-formulated, and then answered..


