# Can we get $f_0,f_1$ uniquely determined by $f$ under the condition $Hom_A(M,Q_0)=0$?

Let $A$ be a finite dimensional algebra over field K. $M,N$ are $A$-modules such that $0 \rightarrow P_1 \rightarrow P_0 \rightarrow M \rightarrow 0$ and $0 \rightarrow Q_1 \rightarrow Q_0 \rightarrow N \rightarrow 0$ are minimal projective resolutions of them, respectively. Let $f: M \rightarrow N$ be an morphism. Then by the projectivity of $P_1,P_0$, we have $f_1,f_0$ such that the following diagram commutes:

Also we know that $f_1,f_0$ may not be unique. I want to know under the condition that $Hom_A(M,Q_0)=0$, can we get $f_0,f_1$ uniquely determined by $f$?

The condition $Hom_A(M,Q_0)=0$ is not sufficient to ensure that $f_0$ and $f_1$ are uniquely determined by $f$.
To see a counter-example, let $Q=1\to 2\to 3$ be a quiver, $A = KQ$ be its path algebra, and let $M$ be the simple (left) $A$-module at vertex $2$ and $N$ be the simple module at vertex $1$. The projective resolutions of $M$ and $N$ look like this: $$0\to P(3) \to P(2) \to M\to 0 \quad \textrm{and} \quad 0\to P(2) \to P(1) \to N\to 0.$$ Now, if we take $f:M\to N$ to be the zero morphism, we see that letting $f_1$ be the inclusion of $P(3)$ into $P(2)$ and $f_0$ be the inclusion of $P(2)$ into $P(1)$ yields a commutative diagram as in your question. Since putting $f_0=f_1=0$ also yields such a commutative diagram, we get that $f_1$ and $f_0$ are not uniquely determined by $f=0$, even though $Hom_A(M, P(1)) = 0$.
However, a necessary and sufficient condition for $f_0$ and $f_1$ to be uniquely determined by $f$ would be that $Hom_A(P_0, Q_1)=0$.