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 $u:P \to A$)?
