Maps of complexes are homotopic if and only if a certain diagram exists $\DeclareMathOperator{\mc}{Mc}$ $\DeclareMathOperator{\diff}{Diff}$ $\DeclareMathOperator{\fct}{Fct}$ $\newcommand{\c}{\mathscr{C}}$I'm trying to understand and prove the following statement :
Let $\c$ be an additive category, and $f,g:X\to Y$ be two morphisms in $C(\c)$. Prove that $f,g$ are homotopic if and only if there exists a commutative diagram in $C(\c)$ : $\require{AMScd}$
\begin{CD}
Y@>\alpha(f)>> \mc(f) @> \beta(f)>> X[1]\\
@| @VuVV @|\\
Y@>\alpha(g)>> \mc(g) @> \beta(g)>> X[1]
\end{CD}
To clarify notations (I'm unsure whether it is standard or not) :

*

*A differential object $(X^\bullet, d^\bullet )$ in $\c$ is a sequence of objects of $\c$ $X^k$ and connecting morphisms $d^k:X^k\to X^{k+1}$, $k\in \mathbb{Z}$, i.e. a functor $F:\mathbb{Z}\to \c$. Hence the category of differential objects in $\c$ $\diff(\c)$ is $\fct(\mathbb{Z},\c)$, which is additive.

*A complex is a differential object such that $d^kd^{k-1}=0$ for all $k\in \mathbb{Z}$

*$C(\c)$ is the full additive subcategory of $\diff(\c)$ consisting of complexes.

*For $f:X\to Y$ in $C(\c)$ we define the mapping cone of $f$ as $\mc(f)^n = X[1]^n\oplus Y^n=X^{n+1}\oplus Y^n$ with differential $d^n_{\mc(f)} = \bigl(\begin{smallmatrix} -d^{n+1}_X & 0 \\ f^{n+1} & d^n_Y \end{smallmatrix}\bigr)$.

*The maps $\alpha(f)$ and $\beta(f)$ are the natural maps $Y\to \mc(f)$ and $\mc(f) \to X[1]$ respectively, we have $\beta(f)\alpha(f)=0 $.

*$f:X\to Y$ in $C(\c)$ is homotopic to $0$ if for all $n$ there is $s^n:X^n\to Y^{n-1}$ such that $f^n=s^{n+1}d_X^n+d_Y^{n-1}s^n$. Two morphisms $f,g$ are homotopic if $f-g$ is homotopic to $0$.

*Given a complex $X$, the complex $X[p]$ is defined as $X[p]^n=X^{n+p}$ with differential $d_{X[p]}^n = (-)^pd_X^n$
Now to prove that if $f,g$ are homotopic then such a diagram exists, I think I only need to find $u:\mc(f)\to\mc(g)$ making the diagram above commute. I tried to define such a map but I have very little intuition for everything at play here, so I essentially tried to make up something using $f,g,s$ but got nowhere. Since $u$ is a morphism in $C(\c)$ this tells me it should make this diagram commute :
$\require{AMScd}$
\begin{CD}
X[1]^n\oplus Y^n@>d_{\mc(f)}^n>> X[1]^{n+1}\oplus Y^{n+1}\\
@Vu^nVV @Vu^{n+1}VV \\
X[1]^n\oplus Y^n@>d_{\mc(g)}^n>> X[1]^{n+1}\oplus Y^{n+1}
\end{CD}
But this didn't help me get much further.
I had an idea for the reverse implication but I am not comfortable enough with the material to see it through. Assuming I have the map $u$, for each $n$ I need to define a map $X^n \to Y^{n-1}$. The only map I could come up with was the composition $$X^{n}\to X^{n}\oplus Y^{n-1} = \mc(f)^{n-1} \xrightarrow{u^{n-1}}\mc(g)^{n-1} =X^{n}\oplus Y^{n-1} \to Y^{n-1}$$ but I'm not entirely sure this makes sense, and similarly I don't know how to check whether this map would satisfy the definition of a homotopy between $f-g$ and $0$. Any help is appreciated thanks!
 A: I think the easiest way to proceed is just to do a direct computation. We know that $u^n$ is given by some matrix $\begin{pmatrix}a^n & b^n \\ c^n & e^n\end{pmatrix}$, with $a^n:X[1]^n\to X[1]^n$, $b^n:Y^n\to X[1]^n$, $c^n:X[1]^n\to Y^n$, $e^n:Y^n\to Y^n$. Knowing that $u^n$ makes the diagram commute will drastically restrict which kinds of $u^n$ are allowed, but I'm going to do the computation for when this is a chain map first. The full complexity of it makes it clear how important the commuting condition is. This is a map of chain complexes if and only if it commutes with the differential, which you've told us is $\begin{pmatrix}-d^{n+1}_X & 0 \\  f^{n+1} & d^n_Y \end{pmatrix}$. So we need $d^n_{\text{Mc}(g)}\circ u^n = u^{n+1}\circ d^n_{\text{Mc}(f)}$, which in terms of matrices is
$$
\begin{pmatrix}-d^{n+1}_X & 0 \\  g^{n+1} & d^n_Y \end{pmatrix}
\begin{pmatrix}a^n & b^n \\ c^n & e^n\end{pmatrix}
=
\begin{pmatrix}a^{n+1} & b^{n+1} \\ c^{n+1} & e^{n+1}\end{pmatrix}
\begin{pmatrix}-d^{n+1}_X & 0 \\  f^{n+1} & d^n_Y \end{pmatrix}.
$$
Multiplying this out, we get
$$
\begin{pmatrix}
-d^{n+1}_Xa^n & -d^{n+1}_Xb^n \\  
g^{n+1}a^n+d^n_Yc^n & g^{n+1}b^n+d^n_Ye^n
\end{pmatrix}
=
\begin{pmatrix}
-a^{n+1}d^{n+1}_X+b^{n+1}f^{n+1} & b^{n+1}d^n_Y \\  
-c^{n+1}d^{n+1}_X+e^{n+1}f^{n+1} & e^{n+1}d^n_Y 
\end{pmatrix}.
$$
Extracting the individual equations, we get
$$-d^{n+1}_Xa^n=-a^{n+1}d^{n+1}_X+b^{n+1}f^{n+1},$$
$$-d^{n+1}_Xb^n=b^{n+1}d^n_Y,$$
$$g^{n+1}a^n+d^n_Yc^n = -c^{n+1}d^{n+1}_X+e^{n+1}f^{n+1},$$
$$g^{n+1}b^n+d^n_Ye^n=e^{n+1}d^n_Y.$$
The second equation says that $b$ is a chain map $Y\to X[1]$. The others don't quite tell us anything interesting yet, but we still haven't factored in the requirement that $u$ makes the diagram commute.
$\alpha(f)$ is given by the matrix $\begin{pmatrix}0 \\ 1\end{pmatrix}$, and
$\beta(f)$ by $\begin{pmatrix} 1 & 0 \end{pmatrix}$.
So from $\alpha(g) = u \circ \alpha(f)$, we get that
$b^n=0$, $e^n=1$, and from $\beta(g)\circ u = \beta(f)$, we get that $a^n=1$, and $b^n=0$. Thus factoring in the commuting conditions, we have that
$u^n=\begin{pmatrix}1&0\\c^n&1\end{pmatrix}$ for some map $c^n:X[1]^n\to Y$ subject to the conditions
$$-d^{n+1}_X = -d^{n+1}_X+0,$$
$$0=0,$$
$$g^{n+1}+d^n_Yc^n = -c^nd^{n+1}_X+f^{n+1},$$
$$0+d^n_Y=d^n_Y.$$
In other words, all the conditions for $u$ to be a chain map become trivial  except for the third, which when rearranged is
$$f^{n+1}-g^{n+1} = d^n_Yc^n + c^nd^{n+1}_X,$$
which says that $c$ is a chain homotopy between $f$ and $g$.
In other words, in order for the commuting condition to be true, the $u^n$s  must all be of the form
$$\begin{pmatrix} 1 & 0 \\ c^n & 1 \end{pmatrix},$$
and if $u$ is of this form, it is a chain map if and only if the $c^n$s give a chain homotopy between $f$ and $g$.
