The Stacks project

Definition 59.70.1. Let $j : U \to X$ be an étale morphism of schemes.

  1. The restriction functor $j^{-1} : \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale}) \to \mathop{\mathit{Sh}}\nolimits (U_{\acute{e}tale})$ has a left adjoint $j_!^{Sh} : \mathop{\mathit{Sh}}\nolimits (U_{\acute{e}tale}) \to \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale})$.

  2. The restriction functor $j^{-1} : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}(U_{\acute{e}tale})$ has a left adjoint which is denoted $j_! : \textit{Ab}(U_{\acute{e}tale}) \to \textit{Ab}(X_{\acute{e}tale})$ and called extension by zero.

  3. Let $\Lambda $ be a ring. The restriction functor $j^{-1} : \textit{Mod}(X_{\acute{e}tale}, \Lambda ) \to \textit{Mod}(U_{\acute{e}tale}, \Lambda )$ has a left adjoint which is denoted $j_! : \textit{Mod}(U_{\acute{e}tale}, \Lambda ) \to \textit{Mod}(X_{\acute{e}tale}, \Lambda )$ and called extension by zero.

Comments (4)

Comment #74 by Keenan Kidwell on

Is "...functor is right exact, so it has a left adjoint..." what is intended? Is there some result elsewhere which states that, in this context, having a left adjoint is implied by being right exact?

Comment #81 by on

Fixed. There is such a result (maybe with more hypotheses), but since we have an explicit description of the extension by zero functor it is better not to appeal to it here. Thanks!

Comment #3538 by Timo Keller on

In (1), the functor goes in the wrong direction.

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View Definition 59.70.1 as pdf