Consider a cohomological delta functor $T^*:\mathscr{A}\to\mathscr{B}$ between abelian categories such that $T^i$ is an effaceable functor for all $i>0$, i.e. for all $i>0$, and for any $A\in\mathscr{A}$, there exists an injection $u_A:A\hookrightarrow M_A$ in $\mathscr{A}$ such that $T^i(u_A) = 0$. Grothendieck famously proved in his Tōhoku paper that this implies that $T^*$ is a universal $\delta^*$-functor, which he later on used to prove his Vanishing Theorem.

However, in his Tōhoku paper, Grothendieck introduces a tremendous amount of unique terminology and definitions, a bit different from other authors of homological algebra (e.g. Weibel), so I've been having a bit of trouble following both the proof itself, and his work up to that proof, so I was wondering if there is an easier proof.

First, let's fix some $\delta^*$-functor $S:\mathscr{A}\to\mathscr{B}$, and suppose we are given $f^0:T^0\to S^0$. My general idea is to go by induction (as I see that Grothendieck does), and assume that for some $n\ge 1$, we have defined $f^i:T^i\to S^i$ for all $0\le i<n$ commuting with all connecting morphisms. Then it is sufficient to show that we can uniquely define $f^n:T^n\to S^n$ commuting with the connecting morphisms, i.e. for any exact sequence $$0\to A\to B\to C\to 0$$ in $\mathscr{A}$, the following square commutes: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} T^{n-1}(C) & \ra{\delta^n} & T^n(A)\\ \da{f^{n-1}_C} & & \da{f^n_A}\\ S^{n-1}(C) & \ras{\delta^n} & S^n(A) \\ \end{array} $$

Now, my first thoughts are to fix the maps $u_A$, and use them to define a short exact sequence $$0\to A\xrightarrow{u_A}M_A\xrightarrow{p_A}P_A\to 0$$for all $A\in\mathscr{A}$, then if we let this is exact sequence induce the connecting maps ${^T\!\!\delta_A^n}:T^{n-1}(C)\to T^n(A)$ and ${^S\!\!\delta_A^n}:S^{n-1}(C)\to S^n(A)$, however, all this gives me is that ${^T\!\!\delta_A^n}$ is surjective. Is this enough to define $f_A^n$?

Consider a cohomological delta functor $T^*:\mathscr{A}\to\mathscr{B}$ between abelian categories such that $T^i$ is an effaceable functor for all $i>0$, i.e. for all $i>0$, and for any $A\in\mathscr{A}$, there exists an injection $u_A:A\hookrightarrow M_A$ in $\mathscr{A}$ such that $T^i(u_A) = 0$. Grothendieck famously proved in his Tōhoku paper that this implies that $T^*$ is a universal $\delta^*$-functor, which he later on used to prove his Vanishing Theorem.

However, in his Tōhoku paper, Grothendieck skips over most of the fine details of the proof, and it's those details that I'm trying to fill in for myself. Here is my work so far:

Fix some $\delta^*$-functor $S:\mathscr{A}\to\mathscr{B}$, and suppose we are given $f^0:T^0\to S^0$. My general idea is to go by induction (as I see that Grothendieck does), and assume that for some $n\ge 1$, we have defined $f^i:T^i\to S^i$ for all $0\le i<n$ commuting with all connecting morphisms. Then it is sufficient to show that we can uniquely define $f^n:T^n\to S^n$ commuting with the connecting morphisms, i.e. such that for any exact sequence $$0\to A\to B\to C\to 0$$ in $\mathscr{A}$, the following square commutes: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} T^{n-1}(C) & \ra{\delta^n} & T^n(A)\\ \da{f^{n-1}_C} & & \da{f^n_A}\\ S^{n-1}(C) & \ras{\delta^n} & S^n(A) \\ \end{array} $$

Now, my first thoughts are to fix the maps $u_A$, and use them to define a short exact sequence $$0\to A\xrightarrow{u_A}M_A\xrightarrow{p_A}P_A\to 0$$for all $A\in\mathscr{A}$. If we let this exact sequence induce the connecting maps ${^T\!\!\delta_A^n}:T^{n-1}(C)\to T^n(A)$ and ${^S\!\!\delta_A^n}:S^{n-1}(C)\to S^n(A)$, then basic diagram chasing gives us that there exists a unique $f^n_A:T^n(A)\to S^n(A)$ such that the following diagram commutes:

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} T^{n-1}(M_A) & \ra{T^{n-1}(p_A)} & T^{n-1}(P_A) & \ra{{^T\!\!\delta_A^n}} & T^n(A) & \ra{0} & T^n(M_A)\\ \da{f^{n-1}_{M_A}} & & \da{f^{n-1}_{P_A}} & & \da{f^n_A}\\ S^{n-1}(M_A) & \ra{S^{n-1}(p_A)} & S^{n-1}(P_A) & \ras{{^S\!\!\delta_A^n}} & S^n(A) \\ \end{array} $$

which of course gives us uniqueness. But how do we show naturality, and how do we show that the natural transformation $f^n:T^n\to S^n$ (once we've proved naturality) commutes with the connecting morphisms?