How to define a natural map from $H^i(Y,\pi_*\mathscr F)\to H^i(X,\mathscr F)$?

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I'm working through the chapter of Vakil's FOAG about cohomology, and this is one of the following properties he claims:

Let $\pi:X\to Y$ be a quasicompact, separable morphism where $Y$ is a quasicompact, separable $A$-scheme. If $\mathscr F$ is a quasicoherent sheaf on $X$ (so $\pi_*\mathscr F$ is quasicoherent as well), we get a natural map $H^i(Y,\pi_*\mathscr F)\to H^i(X,\mathscr F)$.

He leaves this as an exercise for the reader, and I'm trying to prove it. The natural thing to do would be to define a chain map between the complexes we use to calculate the cohomologies. So if $\{U_1,\dots,U_n\}$ is an affine cover of $Y$, letting $U_I=\cap_{i\in I}U_i$ for $I\subseteq[n]$, we can write $\pi^{-1}(U_i)$ as a finite union $V_{i1}\cup\cdots\cup V_{im_i}$ of affine schemes. Therefore if we let $S=\{(i,j):1\le i\le n,1\le j\le m_i\}$, then we see that $\{V_{ij}:(i,j)\in S\}$ is an affine open cover of $X$. The complexes have the form

$$0\to\prod_{|I|=1\\ I\subseteq[n]}\pi_*\mathscr F(U_I)\to\prod_{|I|=2\\ I\subseteq[n]}\pi_*\mathscr F(U_I)\to\cdots\\ 0\to\prod_{|J|=1\\ J\subseteq S}\mathscr F(V_J)\to\prod_{|J|=2\\ J\subseteq S}\mathscr F(V_J)\to\cdots$$

However, I don't see a "natural" way to define a chain map here. The differential is defined using the restriction maps coming from $U_{I\cup\{j\}}\hookrightarrow U_I$, but in the case above, $\pi^{-1}(U_I)$ is in general going to be larger than $V_J$ where $|I|=|J|=1$.

Can somebody help see the flaw in my approach, or give some suggestions of a better way to do this?

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You are almost there. Let me just do the case where $i=0$, the rest works exactly the same.

We want a map from $\prod_{|I|=1\\ I\subseteq[n]}\pi_*\mathscr F(U_I) \to \prod_{|J|=1\\ J\subseteq S}\mathscr F(V_J)$.

Use that $\pi_*\mathscr F(U_i) = \mathscr F(\pi^{-1}(U_i))$ and the fact that for all $j,i$, $V_{ij} \subset \pi^{-1}(U_i)$. This means that we have restriction maps.

Finally, show that these maps indeed come together to give a map of complexes.


A remark: This map exists in great generality, take a look at: http://math.stanford.edu/~conrad/248BPage/handouts/basechange.pdf

You might want to come back to that after you have read a bit more.