On pushouts and mapping cylinders in exact categories Let $\mathcal N$ be an exact category and $C\mathcal N$ be the category of chain complexes with its usual exact structure. We have here the usual notion of "mapping cylinder" of a chain map. If $f:N\to M$ is a chain map, denote by $T(f)$ its mapping cylinder. Denote by $j_1:N\to T(f)$ and $j_2:M\to T(f)$ the inclusion on the first and third factors respectively. These are admissible monomorphisms.
Suppose we have a commutative diagram in $C\mathcal N$:

where $\alpha$ and $\beta$ are admissible monomorphisms.
Let $t:T(f)\to T(f')$ be the induced map; that is, $t=\left(\begin{smallmatrix} \alpha & 0 & 0 \\ 0 & \alpha[-1] & 0 \\ 0 & 0 & \beta  \end{smallmatrix}\right)$.
I want to prove that the induced arrow $s$ in the pushout diagram below is an admissible monomorphism:

Notice that every arrow there but $s$ is an admissible monomorphism.
So I was able to show that $s$ has a cokernel (it's the cokernel $e$ of the map $T(f)\oplus (A'\oplus B') \to T(f')$ given by the matrix $(t \hspace{.6cm} j_1'\oplus j_2')$).
But I'm stuck at showing that $s$ is also the kernel of $e$. I've tried several different things for a couple of days now but I'm just absolutely stuck...
 A: It's probably easiest to compute $P$ and the map $s$ explicitly. It turns out that $P$ has components $P_n = A_{n}' \oplus A_{n-1} \oplus B_{n}'$ and that
$s_n = 1_{A_{n}'} \oplus \alpha_{n-1} \oplus 1_{B_{n}'}$, which is obviously an admissible monomorphism since it is the direct sum of admissible monomorphisms.

Some details, using Weibel's sign conventions on p.20:$\require{AMScd}$
The complex $T(f)$ has components
$A_{n} \oplus A_{n-1} \oplus B_n$ and differential
$\begin{bmatrix}
d^{A}_{n} & 1_{A_{n-1}} & 0 \\
0 & -d^{A}_{n-1} & 0 \\ 
0 & -f_{n-1} & d^{B}_{n}
\end{bmatrix}$, similarly for $T(f')$.
In each degree $n$ there is the push-out diagram
$$\begin{CD}
A_n \oplus B_n
@>{\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1\end{bmatrix}}>> 
A_n \oplus A_{n-1}\oplus B_n
@>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}>>
A_{n-1} \\
@VV{\begin{bmatrix} \alpha_n & 0 \\ 0 & \beta_n\end{bmatrix}}V
@VV{\begin{bmatrix} \alpha_n & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \beta_n \end{bmatrix}}=k_nV @| \\
{A_{n}' \oplus B_{n}'}
@>>{\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1\end{bmatrix}}=q_n>
{A_{n}' \oplus A_{n-1} \oplus B_{n}'}
@>>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}>
A_{n-1}
\end{CD}$$
from which you can see that $P_n = A_{n}' \oplus A_{n-1} \oplus B_{n}'$.
Contemplating the push-out above, you can convince yourself that the differential of $P$ must be given by the matrix
$$
\begin{bmatrix}
d_{n}^{A'} & \alpha_{n-1} & 0 \\
0 & - d_{n-1}^A & 0 \\
0 & -\beta_{n-1} f_{n-1} & d_{n}^{B'}
\end{bmatrix}
$$
and that the induced map $s_n\colon P_n \to T(f')$ is given by
$s_n = 1_{A_{n}'} \oplus \alpha_{n-1} \oplus 1_{B_{n}'}$, as desired.
Note also that this shows that the cokernel of $s$ is equal to the cokernel of $\alpha[-1]$.
