Splitting form of projective presentation vs minimal presentation. Suppose that $A$ is a finite dimensional $k=\bar{k}$-algebra. Consider a finite dimensional left module $M$ and its minimal presentation $$ P^1 \xrightarrow{\pi^1} P^0 \xrightarrow{\pi^0}M \rightarrow 0.$$
Now take another projective presentation $$ Q^1 \xrightarrow{q^1} Q^0 \xrightarrow{q^0}M \rightarrow 0.$$
I've often found in literature that the presentation $Q^\bullet$ has to be isomoprhic as a complex to (for some $\psi$) $$ P^1 \bigoplus \widetilde{Q^1} \xrightarrow{f^1}P^0 \bigoplus \widetilde{Q^0} \xrightarrow{f^0} M \rightarrow 0$$
where $$f^1 = \begin{pmatrix} \pi^1 & 0 \\ 0 & \psi \end{pmatrix}  \hspace{0,5cm} and \hspace{0,5cm} f^0=(\pi^0, 0).$$
I understand why $Q^1$ and $Q^0$ has to split in such a form. It follows from the definition of projective cover. I also understand the form of $f^0$, this is quite obvious, but I don't understand why we can chose the splitting of $Q^1$ and $Q^0$ in a way that the bottom left entry of $f^1$ is zero. For any splitting the upper right entry of $f^1$ has to be zero, but of course this is not true in general for the bottom left entry. Also I don't think that the answer ginen in  Question about minimal projective presentations of a module. is correct. Can you help me please?
 A: In the following, you'll find my idea for showing that we can obtain an isomorphic complex with the diagonal form as desired. I must admit that I was quite surprised to need that much technicality and I am not completely sure I got it all right.
Let us say we obtained an isomorphic complex with the desired direct sum decompositions and with $f^0 = (\pi^1,0)$ as well as
$f^1 = \begin{pmatrix} \pi^1 & 0 \\ \theta & \psi \end{pmatrix}$
for some $\theta : P^1 \to \widetilde{Q^0}$.
From what you said you understood, I think we can agree that this is doable.
We can now consider the diagram
$\require{AMScd}\begin{CD}
P^1 \oplus \widetilde{Q^1} @>{\begin{pmatrix} \pi^1 & 0 \\ \theta & \psi \end{pmatrix}}>> P^0 \oplus \widetilde{Q^0} @>{f^0}>> M \\
@VV{\begin{pmatrix} id & 0 \\ \alpha & id \end{pmatrix}}V @V{id}VV @V{id}VV \\
P^1 \oplus \widetilde{Q^1} @>>{\begin{pmatrix} \pi^1 & 0 \\ 0 & \psi \end{pmatrix}}> P^0 \oplus \widetilde{Q^0} @>{f^0}>> M 
\end{CD}$
and note that it is commutative if and only if
$\psi \circ \alpha = \theta $.
It thus suffices to show that $\theta$ lifts along $\psi$ and for this it suffices to show that $\psi$ is surjective (and then use projectivity of $P^1)$.
In fact, we have $\widetilde{Q^0} \subseteq \ker(f^0) = \text{im}(f^1)$ and so every $x \in \widetilde{Q^0}$ is of the form $x = f^1(a,b)$ for $a \in P^1$ and $b \in \widetilde{Q^1}$. From the direct sum decomposition, we find that $\pi^1(a) = 0$ and so $a \in \ker(\pi^1) = \text{rad}(P^1)$. It follows that $x \in \theta(\text{rad}(P^1)) + \text{im}(\psi)$ so we have $$\widetilde{Q^0} = \theta(\text{rad}(P^1)) + \text{im}(\psi).$$ Since $\theta(\text{rad}(P^1)) \subseteq \text{rad}(\widetilde{Q^0})$ is superfluous, it follows that $\psi$ is surjective and that is what we wanted.
