Let $G$ be a group and $N$ be its normal subgroup. Let $A$ be a $G$-module, then we have the Hochschild-Serre spectral sequence $$ E_2^{p,q} = H^p(G/N, H^q(N,A)) \Rightarrow H^{p+q}(G,A). $$ and $$ E_{p,q}^2 = H_p(G/N, H_q(N,A)) \Rightarrow H_{p+q}(G,A). $$ I'm trying to prove the Hochschild-Serre spectral sequence by applying Grothendieck spectral sequence. For cohomological version, we are considering the composite of functors $$ G-\mathsf{Mod} \xrightarrow{(-)^N} G/N-\mathsf{Mod} \xrightarrow{(-)^{G/N}} \mathsf{Ab} \quad (\star) $$ and for homological version, we are considering $$ G-\mathsf{Mod} \xrightarrow{(-)_N} G/N-\mathsf{Mod} \xrightarrow{(-)_{G/N}} \mathsf{Ab} \quad (\star\star) $$ Then one claims that the composition in $(\star)$ is nothing but $(-)^G$, and composition in $(\star\star)$ is nothing but $(-)_G$.
My Question: Why does this holds?
In Rotman's book, he claimed that $$ \mathrm{Hom}_{\mathbb{Z}(G/N)}(\mathbb{Z}, \mathrm{Hom}_{\mathbb{Z}N}(\mathbb{Z}, A)) = \mathrm{Hom}_{\mathbb{Z}G}(\mathbb{Z}, A) \quad(\dagger) $$ and $$ \mathbb{Z} \otimes_{\mathbb{Z}(G/N)} (\mathbb{Z} \otimes_{\mathbb{Z}N} A) = \mathbb{Z} \otimes_{\mathbb{Z}G} A \quad(\dagger\dagger) , $$ but without proof. I can see that it suffices to prove $(\dagger)$ and $(\dagger\dagger)$, but how to prove $(\dagger)$ and $(\dagger\dagger)$?
P.S.: I have read Weibel's proof on this. He proved this via directly showing the two side inclusions without mentioning $\mathrm{Hom}$ and $\otimes$, but I hope to see a proof via $(\dagger)$ and $(\dagger\dagger)$, which is clearer functorially than Weibel's proof.
Thank you all for commenting and answering!