Suppose that $ f $ and $ f' : C \to D $ are morphisms of chain complexes; Cone($f$) is the mapping cone of $f$; if $f$ and $f'$ are chain homotopic, what is the relation between Cone($f$) and Cone($f'$) ?
[Math] mapping cones of chain homotopic maps
abstract-algebrahomological-algebra
Related Solutions
Let $f:A\to B$ be a chain map. Then, applying the cone construction you have above to the identity chain map $id_A:A\to A$ gives a complex $CA = C(id_A)$ with $CA_n = A_{n-1}\oplus A_n$ and differentials $$d:CA_n\to CA_{n-1},\quad d(x,y) = (-dx,dy-x).$$ What I meant by an extension in the comment above is that there exists a chain map $\bar{f}:CA\to B$ such that $$\bar{f_n}(0,x) = f_n(x),\quad \forall x\in A_n.$$ We then have the following result:
A chain map $f:A\to B$ is null-homotopic if and only if there exists an extension $\bar{f}:CA\to B$ of $f$ to the cone $CA$.
For the direction $(\Rightarrow)$, if $f$ is null-homotopic, there exists a sequence of maps $$\{s_n:A_n\to B_{n+1}\}$$ such that $f_n = d_{n+1}^B\circ s_n+s_{n-1}\circ d_n^A$ for all $n$. Then, we define an extension $\bar{f}:CA\to B$ by: $$\bar{f_n}:CA_n\to B_n,\quad \bar{f_n}(x,y) = f_n(y)-s_{n-1}(x).$$ Showing that this is a chain map is elementary, so I'll leave it as an exercise.
For the direction $(\Leftarrow)$, you can use the condition that $\bar{f}$ is a chain map to build chain homotopies: $$s_{n-1}:A_{n-1}\to B_n,\quad s_{n-1}(a) = -\bar{f_n}(a,0).$$ There's nothing fancy needed to show this gives a chain homotopy, so I'll leave this as an exercise as well.
I can only sketch a solution. First consider what $\tilde{f}$ could be if $f=0$ (not homotopic to zero but just zero). Then what would happen is that the C-E (Cartan Eilenberg) resolution over H and B (the homology and boundary projective resolutions which generate the C-E resolution) would be a chain map over zero. What are the possible chain maps over projective resolutions over zero? They must be chain homotopic to zero. There's a version of horseshoe lemma which allows you to extend these homotopies to the C-E resolution, this gives a vertical homotopy to zero over any identically zero map. Now returning to the question, we wonder what happens if we are a C-E resolution over a homotopy zero map. In this case, simply lift the horizontal homotopy to projective resolutions to get a C-E map which is horizontally homotopic to zero. Now subtract this C-E map from the one given, it must be a C-E map over zero which is vertically homotopic to zero. This shows that any C-E map over a chain map homotopic to zero must be homotopic to zero.
Best Answer
I see this is an old question (possibly a homework), but I found it while googling something else, so let me write down an answer, since I think the comments above are not 100% correct. If you have a chain homotopy between morphisms of complexes $f, f'\colon C_\bullet \to D_\bullet$, then you can use this chain homotopy to construct by hand a (noncanonical) isomorphism of complexes $Cone (f) \cong Cone (f')$.
A chain homotopy between $f$ and $f'$ is a family of morphisms $h_n\colon C_n\to D_{n+1}$ such that
$$\tag{*} f_n - f_n' = d_{n+1}^D \circ h_n + h_{n-1} \circ d_n^C.$$
Now $Cone (f)$ and $Cone (f')$ are built of objects $D_n \oplus C_{n-1}$ and differentials \begin{align*} d_n &= \begin{pmatrix} d_n^D & f_{n-1} \\ 0 & -d_{n-1}^C \end{pmatrix} : D_n \oplus C_{n-1} \to D_{n-1} \oplus C_{n-2} , \\ \quad d_n' &= \begin{pmatrix} d_n^D & f_{n-1}' \\ 0 & -d_{n-1}^C \end{pmatrix} : D_n \oplus C_{n-1} \to D_{n-1} \oplus C_{n-2} , \end{align*} where we are using the matrix notation for linear maps between direct sums (i.e., if $p : X \to Z$, $q : X \to W$, $r : Y \to Z$ and $s : Y \to W$ are any four linear maps, then $\begin{pmatrix} p & q \\ r & s \end{pmatrix}$ denotes the linear map from $X \oplus Y$ to $Z \oplus W$ that acts as $p + q$ on $X$ and acts as $r + s$ on $Y$).
We can define a morphism
$$u_n = \begin{pmatrix} id_{D_n} & h_{n-1} \\ 0 & id_{C_{n-1}} \end{pmatrix}\colon Cone (f) \to Cone (f')$$
It's easy to check that it is a morphism of complexes (multiply the matrices and use the identity (*)). In the other direction, you can define a morphism of complexes
$$v_n = \begin{pmatrix} id_{D_n} & -h_{n-1} \\ 0 & id_{C_{n-1}} \end{pmatrix}\colon Cone (f') \to Cone (f)$$
Now you just multiply matrices to see that $u_\bullet$ and $v_\bullet$ are mutually inverse maps of complexes, QED.
And a similar thing which is used sometimes: if in general, you have a square of morphisms of complexes
$$\require{AMScd} \begin{CD} B_\bullet @>f_\bullet>> C_\bullet\\ @VVu_\bullet V @VVv_\bullet V\\ D_\bullet @>g_\bullet>> E_\bullet \end{CD}$$
that commutes up to homotopy (i.e. there is some chain homotopy $v_\bullet \circ f_\bullet \simeq g_\bullet \circ u_\bullet$), then you can use this chain homotopy to write down a morphism $Cone (f) \to Cone (g)$ that gives you a commutative diagram (with the two squares commuting strictly)
$$\require{AMScd} \begin{CD} 0 @>>> C_\bullet @>>> Cone (f) @>>> B_\bullet [-1] @>>> 0\\ @. @VVv_\bullet V @VV V @VVu_\bullet[-1]V \\ 0 @>>> E_\bullet @>>> Cone (g) @>>> D_\bullet [-1] @>>> 0\\ \end{CD}$$
The formula for this morphism is something like $\begin{pmatrix} v_n & h_{n-1} \\ 0 & u_{n-1} \end{pmatrix}$.