This should a be basic enough question, but I’m a little confused.

In proving that $H^*(X,\mathbf{Q}_{\ell})$ is functorial (in the sense of Weil cohomology theories: see axiom D2 here) as $X$ ranges over smooth projective varieties over an algebraic closure $k$ of a finite field, one argues proving that for any map $f : X\to Y$ of smooth projective varieties over $k$, we have that the natural map

$$\alpha_f : f^{-1}\mathbf{Q}_{\ell}\to \mathbf{Q}_{\ell}$$

in $D_{\rm cons}(X,\mathbf{Q}_{\ell})$, is an isomorphism.

Here the inverse image functor is constructed as $(\varprojlim\ f^{-1}_{\rm ét}\mathbf{Z}/\ell^n\mathbf{Z})[1/\ell]$ so to speak, where $f_{\rm ét}^{-1}$ is the étale inverse image on abelian sheaves.

Why is it so important to know that $\alpha_f$ is an

isomorphism? One always has a map $\alpha_f$, and could define the effect of $f$ on cohomology, $f^* : H^*(Y,\mathbf{Q}_{\ell})\to H^*(X,\mathbf{Q}_{\ell})$, to be the composition:$$H^*(Y,\mathbf{Q}_{\ell})\to H^*(X, f^{-1}\mathbf{Q}_{\ell})\xrightarrow{\alpha_f} H^*(X,\mathbf{Q}_{\ell})$$

where the first map always exists by functoriality of the étale site, without knowing that $\alpha_f$ is an isomorphism.

Where is the fact that $\alpha_f$ is an isomorphism, so crucially used? Is it used in showing that $f^*$ has good properties: namely, it induces a map of graded commutative $\mathbf{Q}_{\ell}$-algebras? Or perhaps on an even more basic level, to show that for a composition $f\circ g$, we have $(f\circ g)^* = g^*\circ f^*$?