This pullback induces a basic sheaf isomorphism in "an Invitation to algebraic geometry" This sheave isomorphism $ \{ \mathcal{O}_{W} \simeq H_{\lambda *}\mathcal{O}_{U_{\lambda}}(U) | \forall U \subset W, g \rightarrow g \circ H_{\lambda} \}$ is supposed to be induced by the pullback technique shown in the first picture. I can tell the elements $g$ and $g \circ H_{\lambda}$ are isomorphic but couldn't there be more elements in $H_{\lambda *}\mathcal{O}_{U_{\lambda}}(U) = \mathcal{O}_{U_{\lambda}}(H^{-}(U))$ than just $g \circ H_{\lambda}$?
I have found the inverse of the morphism from three coordinates to two coordinates returns the initial three coordinates again because it has to be possible to compute the composed functions, so the sheaf of $H^{-}(U \subset W_{\lambda})$ should include functions that aren't necessarily composed with the morphism. Then this isomorphism $\mathcal{O}_{W_\lambda} \simeq \mathcal{O}_{U_{\lambda}}(H^{-}(U))$ is just to a subset of $\mathcal{O}_{U_{\lambda}}(H^{-}(U))$ right?


 A: The trick here is that you are calculating the pullback of a homeomorphism which is an isomorphism as well. Therefore you can take a look at the inverse of this homeomorphism $W_\lambda \overset{H^{-1}_\lambda}{\longrightarrow}W_\lambda$ to get the pullback with its inverse pullback for $W\subset W_\lambda, U\subset U_\lambda$
\begin{align}
H_\lambda^\#: \mathcal{O}_{W_\lambda}(W)&\to \mathcal{O}_{U_\lambda}(H^{-1}(W)) \\
g &\mapsto g \circ H_\lambda \\
\\
H_\lambda^{-1\#}: \mathcal{O}_{U_\lambda}(U)&\to \mathcal{O}_{W_\lambda}(H(U)) \\
g &\mapsto g \circ H_\lambda^{-1} \\
\end{align}
The homeomorphism property is important, otherwise $H(U)$ will not necessarily be open if $U$ is open. You can see easily that $H_\lambda^{-1\#} \circ H_\lambda^{\#} = id_{\mathcal{O}_{W_\lambda}}$ and $H_\lambda^{\#} \circ H_\lambda^{-1\#} = id_{\mathcal{O}_{U_\lambda}}$
In the given example with three and two coordinates, we don't have an isomorphism, as $\mathbb{A}^3 \not\simeq \mathbb{A}^2$, therefore you can't find an inverse pullback morphism.
Or do you mean by the "inverse of the morphism" not the real inverse, but the pullback between their coordinate rings (as this feels like an inverse)?
I hope this is what you're searching for - if not, please tell me what is missing in my explanation
