# Coeffaceable implies universial $\delta$-functor

My question is essentially about Grothendieck's Tohoku paper Proposition 2.2.1 but in the context of coeffaceable instead of effaceable. Grothendieck's paper does not give much suggestions to my questions other than the words standard technique. Let me be more specific.

Let $\mathcal{A}, \mathcal{B}$ be two abelian categories. A functor $F:\mathcal{A} \to \mathcal{B}$ is coeffaceable if for every $A$ there is a surjection $u:P \to A$ such that $F(u) = 0$. If $F_*$ is a homological $\delta$-functor such that each $F_n$ is coeffaceable (except $F_0$), then $F_*$ is universal.

The proof goes by induction. Suppose that $T_*$ is a homological $\delta$-functor and that $\phi_0 : T_0 \to F_0$ is given. We want to show that there is a unique extension of $\phi_0$ to a morphism of $\delta$-functors $\phi:T_* \to F_*$. Suppose inductively that $\phi_i : T_i \to F_i$ are defined for $0 \leq i < n$. Given $A$ in $\mathcal{A}$, select a surjection $u: P \to A$ such that $F_n(u) = 0$. We have a short exact sequence $0 \to K \to P \to A \to 0$ and a commutative diagram

\begin{array} \mbox{} & T_n(A) \stackrel{\delta_n}{\longrightarrow} & T_{n-1}(K) \stackrel{}{\longrightarrow} & T_{n-1}(P) \\ & & \downarrow{\phi_{n-1}(K)} & \downarrow{\phi_{n-1}(P)} \\ F_n(P) \stackrel{F_n(u)=0}{\longrightarrow} & F_n(A) \stackrel{\delta_n}{\longrightarrow}& F_{n-1}(A) \stackrel{}{\longrightarrow} & F_{n-1}(P) \end{array}

As $F_n(u)=0$, $\delta_n$ on the second row is injective. This implies that there exists a map $\phi_n(A):T_n(A) \to F_n(A)$.

The question I have is to show that $\phi_n(A)$ is defined independent of choice of $u : P \to A$. The idea suggested in Weibel's book (Exercise 2.4.5) goes as follow:

If there is another $u' : P' \to A$ such that $F_n(u') = 0$, there we have

\begin{array} \mbox{} 0 \stackrel{}{\longrightarrow} & K' \stackrel{}{\longrightarrow} & P' \stackrel{}{\longrightarrow} & A \stackrel{}{\longrightarrow} & 0\\ & & & \downarrow{\mathrm{id}} &\\ \mbox{} 0 \stackrel{}{\longrightarrow} & K \stackrel{}{\longrightarrow} & P \stackrel{}{\longrightarrow} & A \stackrel{}{\longrightarrow} & 0\\ \end{array}

What I would like to see is a morphism from $P'$ to $P$, this is true if we assume $P'$ is projective (in the case when $F_n$ are left derived functors and assuming $\mathcal{A}$ has enough projectives). I have had a hard time figuring out why there should be a morphism. If there is not a morphism from $P'$ to $P$, how to proceed to show that $\phi_n(A)$ is well-defined (independent of choice of $u:P \to A$)? Any suggestions is welcome!

Hint: Consider the fiber product $P \times_A P'$ and choose an epimorphism (not surjection) $Q \to P \times_A P'$ which is mapped to $0$ by $F$.