Lemma 6.24.8. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a sheaf of rings on $X$. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. Let $\mathcal{G}$ be a sheaf of $f_*\mathcal{O}$-modules. Then

$\mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(\mathcal{O})}( \mathcal{O} \otimes _{f^{-1}f_*\mathcal{O}} f^{-1}\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}).$

Here we use Lemmas 6.24.6 and 6.24.5, and we use the canonical map $f^{-1}f_*\mathcal{O} \to \mathcal{O}$ in the definition of the tensor product.

Proof. This follows from the equalities

\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(\mathcal{O})}( \mathcal{O} \otimes _{f^{-1}f_*\mathcal{O}} f^{-1}\mathcal{G}, \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(f^{-1}f_*\mathcal{O})}( f^{-1}\mathcal{G}, \mathcal{F}_{f^{-1}f_*\mathcal{O}}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}). \end{eqnarray*}

which are a combination of Lemma 6.20.2 and 6.24.7. $\square$

## Comments (0)

There are also:

• 4 comment(s) on Section 6.24: Continuous maps and sheaves of modules

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 008Z. Beware of the difference between the letter 'O' and the digit '0'.