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

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!

• It is really tortuous having to distinguish $\mathscr I$ and $\mathscr J$ here when scanning the text. – Pedro Tamaroff Aug 12 '16 at 8:19
• How are you composing $\varepsilon$ and $\varphi$ if they are not composable? According to what you write $J^*$ and $I^*$ live in different categories. – Pedro Tamaroff Aug 12 '16 at 8:31
• @PedroTamaroff I have replaced \mathscr with \mathcal, added two diagrams and fixed typos. Hope it's clearer now. – Yai0Phah Aug 12 '16 at 13:55
• The existence of such comparison morphism should be guaranteed by the formal homological properties of injective resolutions and injective objects. You no longer get a commutative diagram (like when one lifts morphisms between complexes concentrated in degree zero) because you are resolving complexes themselves, but apart from this you should be able to be patient enough to construct the map yourself. It is a bit tortuous but you should start at the level $0$, and inductively construct it. It is not clear to me why you're worrying about $\phi$ being a quasi-isomorphism. – Pedro Tamaroff Aug 12 '16 at 20:26
• (N.B.: I am not claiming this is easy, but it is doable, and should not depend on the context you're working in. This seems one of the reasons the author omits such construction.) – Pedro Tamaroff Aug 12 '16 at 20:27

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.