Group Cohomology: Change of Groups Let $f:G'\to G$ be a group homomorphism and let $A$ be a $G$-module.
We regard $A$ as a $G'$-module via restriction of scalars along $f$.
I am struggling to understand the following statement:
"$f$ induces group homomorphisms $H^q(G,A)\to H^q(G',A)$ for each $q\geq 0.$"

Concrete Interpretation
For each $q\geq 0$ there is a natural group homomorphism
$$\Phi_q:\text{Hom}_G(\mathbb{Z}[G^{q+1}],A)\to\text{Hom}_{G'}(\mathbb{Z}[(G')^{q+1}],A)$$
determined by
$$\Phi_q(\alpha)(g_0',\ldots,g_q')=\alpha(f(g_0'),\ldots,f(g_q')).$$
These induce (as they compatible with the standard resolutions) group homomorphisms
$$H^q(G,A)\to H^q(G',A).$$

Problem
Doesn't this mean the induced map for $q=0$ is just the identity (edit: inclusion, not identity) map?
 A: 
These induce (as they compatible with the standard resolutions) group
  homomorphisms $$H^q (G,A) \to H^q (G',A).$$

Well, actually, these morphisms have nothing to do with a particular resolution, they exist by some abstract nonsense. By definition, group cohomology is given by
$$H^q (G,A) = (R^q (-)^G) (A)$$
—these are the right derived functors of the left exact functor of fixed points
$$(-)^G\colon \mathbb{Z}[G]\textit{-Mod}\to \mathbb{Z}[G]\textit{-Mod}.$$
Now a group homomorphism
$$f\colon G'\to G$$
gives us an exact forgetful functor
$$f^\#\colon \mathbb{Z} [G]\textit{-Mod}\to \mathbb{Z} [G']\textit{-Mod}$$
—any $G$-module may be viewed as a $G'$-module via the action $g'\cdot x = f (g') \cdot x$. For the fixed points, it is easy to see that we get natural monomorphisms
$$A^G \rightarrowtail (f^\# A)^{G'}.$$
Naturality means that for any morphism of $G$-modules $A\to B$ we have the corresponding commutative diagram. That is, we have a natural transformation
$$H^0 (G, -) \cong (-)^G \Rightarrow (f^\# -)^{G'} \cong H^0 (G', f^\# (-)).$$
Now it follows from general properties of right derived functors (as they are universal $\delta$-functors, and $H^0 (G', f^\# (-))$ is a $\delta$-functor) that this uniquely extends to natural transformations in higher degrees
$$H^q (G, -) \Rightarrow H^q (G', f^\# (-)).$$
These are not monomorphisms for $q > 0$, as the case $G' = \{ e \}$ suggests: the trivial group has trivial higher cohomology.
In general, if $T^q$ is a $\delta$-functor, then a natural transformation $R^0 F \Rightarrow T^0$ extends uniquely to natural transformations $R^q F\Rightarrow T^q$ that give you for any short exact sequence $0 \to K\to M \to N\to 0$ a commutative ladder with exact lines

