mapping cones of chain homotopic maps 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($f'$) ?
 A: 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 $\begin{pmatrix}
v_n & h_{n-1} \\
0 & u_{n-1}
\end{pmatrix}$.
