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!
Thanks in advanced!
 A: To fix notation, denote the map $T_n(A) \rightarrow F_n(A)$ defined using the epimorphism $u: P \rightarrow A$ by $\phi^u_n(A)$. The question is to show that $\phi^u_n(A) = \phi^{u'}_n(A)$ for any other epimorphism $u': P' \rightarrow A$ with $F_n(u')=0$. Let $u'': P'' \rightarrow A$ be the fiber product of $u$ and $u'$, and let $K''$ be the kernel of $u''$. Then $u''$ is epic, since $u$ and $u'$ are. Moreover, $F_n(u'')$ is zero since it factorizes via $F_n(u),$ and by assumption $F_n(u)=0$. The projection $P'' \rightarrow P$ induces a map $K'' \rightarrow K$ fitting into the commutative diagram with exact rows
\begin{array}
\mbox{} 0 \stackrel{}{\longrightarrow} & K'' \stackrel{}{\longrightarrow} & P'' \stackrel{}{\longrightarrow} & A \stackrel{}{\longrightarrow} & 0\\
& \downarrow & \downarrow & \downarrow{\mathrm{id}} &\\
\mbox{} 0 \stackrel{}{\longrightarrow} & K \stackrel{}{\longrightarrow} & P \stackrel{}{\longrightarrow} & A \stackrel{}{\longrightarrow} & 0\\
\end{array}
We now have the desired morphism of short exact sequences to which the argument on page 48 of Weibel's book applies. We find that $\phi_n^{u''}(A) = \phi_n^u(A)$, and by symmetry also $\phi_n^{u''}(A) = \phi_n^{u'}(A)$, so that as desired $\phi_n^{u}(A) = \phi_n^{u'}(A)$. In conclusion, the map $\phi_n^u: T_n(A) \rightarrow F_n(A)$ does not depend on the choice of the epimorphism $u: P \rightarrow A$ with $F_n(u)=0$, and may unambiguously be denoted $\phi_n(A)$, omitting the superscript.
To address Mykola's question in a comment to Martin's answer, suppose we have a morphism $f: A' \rightarrow A$. Let $0 \rightarrow K \rightarrow P \stackrel{u}{\rightarrow} A \rightarrow 0$ be the exact sequence chosen to define $\phi_n(A)$.
Since we are now at liberty to choose an epimorphism $u': P' \rightarrow A'$ with $F_n(u')=0$ of our liking, let us select one which by construction comes with a map $g: P' \rightarrow P$ making the right square of the diagram
\begin{array}
\mbox{} 0 \stackrel{}{\longrightarrow} & K' \stackrel{}{\longrightarrow} & P' \stackrel{u'}{\longrightarrow} & A' \stackrel{}{\longrightarrow} & 0\\
& \downarrow{h} & \downarrow{g} & \downarrow{f} &\\
\mbox{} 0 \stackrel{}{\longrightarrow} & K \stackrel{}{\longrightarrow} & P \stackrel{u}{\longrightarrow} & A \stackrel{}{\longrightarrow} & 0\\
\end{array}
commute. As before, $g$ will induce a map $h$ and the argument in Weibel will go through.
To define $P'$, let $Q$ be the fiber product of $u$ and $f$. Let $\pi_P: Q \rightarrow P$ and $\pi_{A'}: Q \rightarrow A'$ be the projections, satisfying $u\pi_{P} = f\pi_{A'}$. Let $v: P' \rightarrow Q$ be some epimorphism with $F_n(v)=0$, and set $g = \pi_Pv$ and $u' = \pi_{A'}v$. The right square commutes as $ug=(u\pi_P)v=(f\pi_{A'})v=u'v,$ and $F_n(u') = F_n(\pi_{A'})F_n(v)=0.$ Finally, $u'$ is epic, since $v$ and $\pi_{A'}$ are such, the latter because it is the pullback of the epimorphism $u$ along $f$.
A: 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 $F$.
