Group Cohomology is a universal $\delta$-Functor Let $G$ be a group, and let $P_\bullet \to \mathbb{Z}$ be a fixed projective resolution of the trivial $G$-module $\mathbb{Z}$. Then we can define the i-th cohomology group $H^i(G,M)$ of $G$ with coefficients in a $G$-module $M$ to be the i-th cohomology of the chain complex $\mathrm{Hom}_G(P_\bullet,M)$.
With this definition I have no problems to find certain connecting morphisms such that $H^*(G,-)$ forms a cohomological $\delta$-functor. However, I don't see a neat way to show it is universal.
I tried to adapt the proof of Weibel's book that left derived functors are universal, but for doing so I need the fact that any module embeds into some injective module. This seems a bit over the top in my opinion (plus it obfuscates how natural transformations of $H^0(G,-)$ extend to $H^*(G,-)$ explicitly). Are there other ways to show universality of $H^*(G,-)$ which avoid injective modules?
 A: I think universality of derived functors is the right way to prove this. Let me try to explain it, in particular the issue with projectives and injectives.

I tried to adapt the proof of Weibel's book that left derived functors
  are universal

There seems to be some confusion between left and right. Group cohomology is given by the right derived functors of the left exact functor of invariants:
$$\tag{*} H^n (G,M) = (R^n (-)^G) (M).$$
To calculate this, one may start from an injective resolution $A \rightarrowtail I^\bullet$ by $\mathbb{Z}G$-modules, and to construct such a resolution we normally use the fact that the category of $R$-modules for any $R$ has enough injectives:

I need the fact that any module embeds into some injective module.

Now to use projective resolutions instead, as you do, one notes that
$$(-)^G \cong \operatorname{Hom}_{\mathbb{Z}G} (\mathbb{Z},-),$$
and therefore
$$H^n (G,M) \cong \underbrace{(R^n \operatorname{Hom}_{\mathbb{Z}G} (\mathbb{Z},-)) (M) \cong (R^n \operatorname{Hom}_{\mathbb{Z}G} (-,M)) (\mathbb{Z})}_{\operatorname{Ext}^n_{\mathbb{Z}G} (\mathbb{Z},M)}.$$
So thanks to the "balancing of Ext" (the second isomorphism), one may indeed start from a projective resolution $P^\bullet \twoheadrightarrow \mathbb{Z}$, and then apply to it the contravariant functor $\operatorname{Hom}_{\mathbb{Z}G} (-,M)$. Why? If we view it as a functor on the opposite category $\mathbb{Z}G\textit{-Mod}^\text{op}$, then the latter is also an abelian category, and, by abstract nonsense


*

*injective objects in $\mathbb{Z}G\textit{-Mod}^\text{op}$ are exactly projective objects in $\mathbb{Z}G\textit{-Mod}$,

*monomorphisms in $\mathbb{Z}G\textit{-Mod}^\text{op}$ are exactly epimorphisms $\mathbb{Z}G\textit{-Mod}$,

*the fact that $\mathbb{Z}G\textit{-Mod}$ has enough projectives translates to $\mathbb{Z}G\textit{-Mod}^\text{op}$ having enough injectives.


Now if you like, you may just assume from the beginning that you're deriving $\operatorname{Hom}_{\mathbb{Z}G} (-,M)$ and use projective resolutions, as you do. This is not the definition of group cohomology, but a posteriori it's the same thing.

Then, the proof of universality of right derived functors boils down to the following result of Grothendieck.

Let $(T^\bullet,\delta^\bullet)$ be a cohomological $\delta$-functor, which is effaceable in the sense that for every $n > 0$ and every
  object $X$ there exists some monomorphism $X\rightarrowtail Y$ such
  that $T^n (Y) = 0$. Then $(T^\bullet,\delta^\bullet)$ is universal.

The proof is not difficult and it's basically in Weibel. It's just some inductive construction that relies on effaceability.
Normally this theorem is applied as follows:


*

*One works with a category with enough injectives, so for any object there is a monomorphism $X\rightarrowtail I$, where $I$ is injective.

*One constructs right derived functors $R^n F$ using injective resolutions and one notes that $(R^n F) (I) = 0$ for $n > 0$ and any injective $I$.
So derived functors are effaceable, and hence universal.

If for some reason you don't like enough injectives and injective resolutions, recall that in your case we may derive the contravariant functor $\operatorname{Hom}_{\mathbb{Z}G} (-,M)$. So again, you may use instead the existence of enough projectives in $\mathbb{Z}G\textit{-Mod}$.

Here's a sketch of the proof of universality of effaceable functors, as requested. Let me deal with the homological case, for example. We have to see that if $(E_\bullet, \delta_\bullet)$ is an effaceable $\delta$-functor, then it's universal, which means that for any other homological $\delta$-functor $(T_\bullet, \partial_\bullet)$ a natural transformation $f_0\colon T_0\Rightarrow E_0$ uniquely extends to a family of natural transformations $f_n\colon T_n\Rightarrow E_n$ that commute with $\delta_n$ and $\partial_n$. The construction is inductive: assuming that we have $f_{n-1}$, we construct $f_n$.
Let's define $f_{n,M}\colon T_n (M) \to E_n (M)$ for an arbitrary object $M$. As $E_\bullet$ is effaceable by our assumption, there is an epimorphism $N\twoheadrightarrow M$ such that $E_n (N) = 0$. Let $K$ be its kernel:
$$0 \to K \rightarrowtail N \twoheadrightarrow M \to 0$$
As $T_\bullet$ and $E_\bullet$ are $\delta$-functors, this short exact sequence gives us the following long exact sequences:

Here the vertical arrows exist by the induction hypothesis. We have $E_n (N) = 0$, and therefore $E_n (M)$ is the kernel of $E_{n-1} (K) \to E_{n-1} (N)$. We see that the dashed arrow $f_{n,M}\colon T_n (M)\to E_n (M)$ exists and it's unique by the universal property of kernels.
A priori this arrow $f_n$ depends on our choice of $N\twoheadrightarrow M$, so we have to check that different choices lead to the same result. Let $N_1\twoheadrightarrow M$ and $N_2\twoheadrightarrow M$ be two different epimorphisms such that $E_n (N_1) = E_n (N_2) = 0$ and let $K_1$, $K_2$ be their respective kernels. Consider $N = N_1\oplus N_2$ and the canonical inclusions $N_i\to N$. The following commutative diagram

gives us a cube

Its top and bottom face commute because $T_\bullet$ and $E_\bullet$ are $\delta$-functors, its back and front face commute by our construction, and its right face commutes by the induction hypothesis. So we see that $$\delta_n\circ f_{n,M}^{K_i} = \delta_n\circ f_{n,M}^K.$$
But $E_n (N) \cong E_n (N_1)\oplus E_n (N_2) = 0$, so $\delta_n\colon E_n (M)\to E_{n-1} (K)$ is a monomorphism, and the above identity implies
$$f_{n,M}^{K_1} = f_{n,M}^{K_2} = f_{n,M}^K.$$
So our construction indeed does not depend on the choice of $N\twoheadrightarrow M$.
The naturality of $f_{n,M}\colon T_n (M)\to E_n (M)$ in $M$ is verified similarly. Given a morphism $\phi\colon M_1\to M_2$, we consider a diagram with exact rows

It induces a cube

We know that all the faces commute, except for the left face, whose commutativity we are trying to prove (the right face commutes by the induction hypothesis).
But commutativity of the other faces is enough to see that
$$\delta_n \circ f_{n,M_2} \circ T_n (\phi) = \delta_n \circ E_n (\phi) \circ f_{n,M_1}.$$
Proof:

And we are done, again becase $\delta_n$ is a monomorphism.
(Exercise: dualize this proof for cohomological $\delta$-functors.)
