The Stacks project

Proof. With $f : (X, \mathcal{O}_ X) \to (pt, A)$ as above we have $Rf_*(-) = R\Gamma (X, -)$ and our map $\mu $ is adjoint to the map

\[ Lf^*(Rf_*K \otimes _ A^\mathbf {L} Rf_*M) = Lf^*Rf_*K \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*Rf_*M \xrightarrow {\epsilon _ K \otimes \epsilon _ M} K \otimes _{\mathcal{O}_ X}^\mathbf {L} M \]

where $\epsilon $ is the counit of the adjunction between $Lf^*$ and $Rf_*$. If we think of $\xi $ and $\eta $ as maps $\xi : A[-i] \to R\Gamma (X, K)$ and $\eta : A[-j] \to R\Gamma (X, M)$, then the tensor $\xi \otimes \eta $ corresponds to the map1

\[ A[-i - j] = A[-i] \otimes _ A^\mathbf {L} A[-j] \xrightarrow {\xi \otimes \eta } R\Gamma (X, K) \otimes _ A^\mathbf {L} R\Gamma (X, M) \]

By definition the cup product $\xi \cup \eta $ is the map $A[-i - j] \to R\Gamma (X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)$ which is adjoint to

\[ (\epsilon _ K \otimes \epsilon _ M) \circ Lf^*(\xi \otimes \eta ) = (\epsilon _ K \circ Lf^*\xi ) \otimes (\epsilon _ M \circ Lf^*\eta ) \]

However, it is easy to see that $\epsilon _ K \circ Lf^*\xi = \tilde\xi $ and $\epsilon _ M \circ Lf^*\eta = \tilde\eta $. We conclude that $\widetilde{\xi \cup \eta } = \tilde\xi \otimes \tilde\eta $ which means we have the desired agreement. $\square$

[1] There is a sign hidden here, namely, the equality is defined by the composition
\[ A[-i - j] \to (A \otimes _ A^\mathbf {L} A)[-i - j] \to A[-i] \otimes _ A^\mathbf {L} A[-j] \]
where in the second step we use the identification of More on Algebra, Item (7) which uses a sign in principle. Except, in this case the sign is $+1$ by our convention and even if it wasn't $+1$ it wouldn't matter since we used the same sign in the identification $\mathcal{O}_ X[-i - j] = \mathcal{O}_ X[-i] \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ X[-j]$.

Comments (0)

There are also:

  • 2 comment(s) on Section 20.31: Cup product

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Your email address will not be published.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 0FP2. Beware of the difference between the letter 'O' and the digit '0'.