Iversen's Cohomology of Sheaves, pull back maps for sheaf cohomology

Question

In Iversen's Cohomology of Sheaves, the cohomology of sheaf (of abelian groups, the category of which has enough injectives) is defined as follows:

Given a sheaf $\mathcal F$ over $X$, we define cohomology groups $H^\bullet(X,\mathcal F)$ as derived functors $R^\bullet\Gamma(X,\mathcal F)$. If $\mathcal F\to\mathcal I^\bullet$ is an injective resolution, then $H^\bullet(X,\mathcal F)$ is the cohomology of the complex $\Gamma(X,\mathcal I^\bullet)$.

Now given a continuous map $f\colon X\to Y$, a sheaf $\mathcal G$ on $Y$, the author wants to introduce a natural map (or pull back) $f^*\colon H^\bullet(Y,\mathcal G)\to H^\bullet(X,f^*\mathcal G)$. We first choose an injective resolution $\mathcal G\to\mathcal J^\bullet$ on $Y$ and an injective resolution $f^*\mathcal J^\bullet\to\mathcal I^\bullet$ (an injective resolution of a complex is a quasi-isomorphism to a complex of injective objects). Since $f^*$ is exact, $\mathcal I^\bullet$ is an injective resolution of $f^*\mathcal G$. The composite $$\Gamma(Y,\mathcal J)\xrightarrow a\Gamma(X,f^*\mathcal J^\bullet)\to\Gamma(X,\mathcal I^\bullet)$$ gives a morphism $f^*$, where $a$ is the natural adjunction map. One needs to show that this doesn't depend on the choice of resolutions, and moreover one can replace injective resolutions to acyclic resolutions, owing to the following theorem:

(Scolium 5.2 of Iversen's Cohomology of Sheaves, page 100) Given a $\Gamma(Y,-)$-acyclic resolution $t\colon\mathcal G\to\mathcal T^\bullet$ and a $\Gamma(X,-)$-acyclic resolution $s\colon f^*\mathcal G\to\mathcal S^\bullet$ and a morphism of complexes $\psi\colon\mathcal T^\bullet\to f_*\mathcal S^\bullet$ such that $\psi\circ t=f_*s\circ a$, where $a\colon\mathcal G\to f_*f^*\mathcal G$ is the adjunction map of adjoint operators $f_*,f^*$. $$\require{AMScd} \begin{CD} \mathcal G@>t>>\mathcal T^\bullet\\ @VVaV@VV\psi V\\ f_*f^*\mathcal G@>f_*s>>f_*\mathcal S^\bullet \end{CD} $$ Then $f^*\colon H^\bullet(Y,\mathcal G)\to H^\bullet(X,f^*\mathcal G)$ is represented by the chain map $$\Gamma(Y,\mathcal T^\bullet)\xrightarrow\psi\Gamma(Y,f_*\mathcal S^\bullet)\to\Gamma(X,\mathcal S^\bullet)$$

In the proof, he starts by choosing arbitrary injective resolutions $\mathcal T^\bullet\xrightarrow{i_1}\mathcal J^\bullet$ and $f^*\mathcal J^\bullet\xrightarrow{i_2}\mathcal I^\bullet$ and he claims that:

There exists a quasi-isomorphism $\epsilon\colon\mathcal S^\bullet\to\mathcal I^\bullet$ such that $\epsilon\circ\phi\colon f^*\mathcal T^\bullet\to\mathcal I^\bullet$ is homotopic to $i_2\circ f^*i_1$, where $\phi\colon f^*\mathcal T^\bullet\to\mathcal S^\bullet$ is the image of $\psi\colon\mathcal T^\bullet\to f_*\mathcal S^\bullet$ under the Hom-set isomorphism of adjoint operators, namely, the existence of the following homotopy commutative diagram:

$$\require{AMScd} \begin{CD} f^*\mathcal T^\bullet@>\phi>>\mathcal S^\bullet\\ @VVV@V\exists VV\\ f^*\mathcal J^\bullet@>>>\mathcal I^\bullet \end{CD} $$

I don't understand this, since it seems to me that $\phi$ is not generally a quasi-isomorphism (or I cannot see any reason), and the existence is quite unclear for me.

It is clear that the existence of the second homotopy commutative diagram should be deduced from the first commutative diagram. However, I don't know how to properly take advantage of the first diagram, translating things from $\phi$ to $\psi$ to deduce anything. I need help on this.

Any ideas? Thanks!

Answer 1

After discussion with others, I realize how to tackle it:

We first note that the Hom-set adjunction \begin{align*} \operatorname{Hom}(f^*\mathcal G,\mathcal S^\bullet)&\xrightarrow\sim\operatorname{Hom}(\mathcal G,f_*\mathcal S^\bullet)\\ s&\mapsto s'=f_*s\circ a \end{align*}

By naturality, we have the following diagram: $$\require{AMScd} \begin{CD} \operatorname{Hom}(\mathcal T^\bullet,f_*\mathcal S^\bullet)@>{t^*}>>\operatorname{Hom}(\mathcal G,f_*\mathcal S^\bullet)\\ @V\mathrm{adj}VV@V\mathrm{adj}VV\\ \operatorname{Hom}(f^*\mathcal T^\bullet,\mathcal S^\bullet)@>{(f^*t)^*}>>\operatorname{Hom}(f^*\mathcal G,\mathcal S^\bullet) \end{CD}$$ Hence $s=\phi\circ f^*t$, and by exactness of $f^*$, $\phi$ is a quasi-isomorphism. We now deduce from Whitehead's theorem the existence of the homotopy commutative diagram.