Clarification of Theorem 2.4.3 in Weibel (left derived functors are universal $\delta$-functors)

Question

I am a little confused by the last two lines of Weibel's proof. Note that Weibel defines a $\delta$-functor $W_*$ to be universal when incoming morphisms $T_* \to W_*$ are completely characterized by $T_0 \Rightarrow W_0$.

The $\delta$-functor we wish to prove is universal is the left derived $\delta$-functor of a right exact functor $F$ from a category with enough projectives. Weibel defines natural transformations $\phi_n \colon T_n \Rightarrow L_n F$ and shows they are unique. Then he gets to the last bit where he needs to show that, given a short exact sequence, the relevant ladder commutes.

Finally, we need to verify that $\varphi_n$ commutes with $\delta_n$. Given a short exact sequence $0 \to A' \to A \to A'' \to 0$ and a chosen exact sequence $0 \to K'' \to P'' \to A'' \to 0$ with $P''$ projective, we can construct maps $f$ and $g$ making the diagram $$ \require{AMScd} \begin{CD} 0 @>>> K'' @>>> P'' @>>> A'' @>>> 0 \\ @. @VV g V @VV f V @| @. \\ 0 @>>> A' @>>> A @>>> A'' @>>> 0 \end{CD} $$ commute. This yields a commutative diagram $$ \begin{CD} T_n(A'') @> \delta >> T_{n - 1}(K'') @> T(g) >> T_{n - 1}(A') \\ @V \varphi_n VV @VV \varphi_{n - 1} V @VV \varphi_{n - 1} V \\ L_n F(A'') @> \delta >> L_{n - 1} F(K'') @> L_n F(g) >> L_{n - 1} F(A'). \end{CD} $$ Since the horizontal composites are the $\delta_n$ maps of the bottom row, this implies the desired commutativity relation.

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I understand why the squares at the end commute but not why $T(g) \circ \delta$ and $LF(g) \circ \delta$ are the desired maps.

Answer 1

This follows from the following axiom of a homological $\delta$-functor (since you're asking about Weibel, I present you the definition from Weibel):

2. For each morphism of short exact sequences from $0 \to A' \to B' \to C' \to 0$ to $0 \to A \to B \to C \to 0$, the $\delta$’s give a commutative diagram $$ \require{AMScd} \begin{CD} T_n(C') @> \delta >> T_{n - 1}(A') \\ @VVV @VVV \\ T_n(C) @> \delta >> T_{n - 1}(A) \end{CD} \qquad\text{resp.}\qquad \begin{CD} T^n(C') @> \delta >> T^{n + 1}(A') \\ @VVV @VVV \\ T^n(C) @> \delta >> T^{n + 1}(A). \end{CD} $$

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Note that, by functoriality, $T_n(1_{A''}) = 1_{T_n(A'')}$ and $L_nF(1_{A''}) = 1_{L_nF(A'')}$