most recent 30 from http://mathoverflow.net 2013-05-21T22:08:40Z http://mathoverflow.net/feeds/question/61641 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61641/inverse-image-as-the-left-adjoint-to-pushforward Daniel Barter 2011-04-14T03:40:18Z 2011-04-14T04:32:25Z

<p>This is a <a href="http://math.stackexchange.com/questions/32868/inverse-image-as-the-left-adjoint-to-pushforward" rel="nofollow">repost</a> of a question on Math stackexchange. No one is biting at it there, so I guess it is harder than I thought. </p> <p>Assume $X$ and $Y$ are topological spaces, $f : X \to Y$ is a continuous map. Let ${\bf Sh}(X)$, ${\bf Sh}(Y)$ be the category of sheaves on $X$ and $Y$ respectively. Modulo existence issues we can define the inverse image functor $f^{-1} : {\bf Sh}(Y) \to {\bf Sh}(X)$ to be the left adjoint to the push forward functor $f_{*} : {\bf Sh}(X) \to {\bf Sh}(Y)$ which is easily described.</p> <p>My question is this: Using this definition of the inverse image functor, how can I show (without explicitly constructing the functor) that it respects stalks? i.e is there a completely categorical reason why the left adjoint to the push forward functor respects stalks? </p>

http://mathoverflow.net/questions/61641/inverse-image-as-the-left-adjoint-to-pushforward/61647#61647 Todd Trimble 2011-04-14T04:32:25Z 2011-04-14T04:32:25Z

<p>Easy: the stalk at a point $x: 1 \to X$ is a functor $\text{Sh}(X) \to Set$ that may be identified with the inverse image functor </p> <p>$$x^\ast: \text{Sh}(X) \to \text{Sh}(1).$$ </p> <p>Since we have $x^\ast \circ f^\ast \cong (f \circ x)^\ast = (f(x))^\ast$, the inverse image pulls back stalk functors to stalk functors. </p>