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Let $\bf Top$ a convenient category of topological spaces, $G$ a group in $\bf Top$, ${}^G\bf Top$ the category of (left) $G$-spaces, and $Sgrp(G)$ the poset of subgroups of $G$.

Define two functors [and prepare yourself to a couple of slight abuses of notation]:

  1. ${}^G{\bf Top}\times Sgrp(G)^\text{op}\to {\bf Top}\colon (X,H)\mapsto X^H$, sending a space into the subspace of fixed point of the $H$ action on $X$;
  2. $Sgrp(G)\to \mathbf{Top}\colon H\mapsto G/H$ with the induced topology.

Now fix a space $X\in \bf Top$ and consider the functor $Sgrp(G)^\text{op}\times Sgrp(G)\to \bf Top$ defined by $(H,K)\mapsto X^H \times G/K$.

How can you prove (if it is true) that $$ \int^{H\in Sgrp(G)} X^H\times G/H\cong X? $$ purely coendy proofs are welcome.

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  • $\begingroup$ Did you mean for $X$ to be a left $G$-space (you said "fix a space $X \in \mathbf{Top}$", but then refer to $X^H$)? $\endgroup$
    – Todd Trimble
    Jul 21 '13 at 11:45
  • $\begingroup$ You're right: I meant "fix X in ${}^G\bf Top$, and sometimes consider the image of the objects under the obvious forgetful functor". $\endgroup$
    – fosco
    Jul 21 '13 at 11:51
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It doesn't seem to be true. Suppose we take $G = \{-1, 1\}$ with the discrete topology, acting on $X = \mathbb{R}$ by usual multiplication. Here $X^G$ consists of a single point $0$. The coend amounts to a pushout in left $G$-spaces of the diagram

$$\{-1, 1\} \times \mathbb{R} \stackrel{id \times 0}{\leftarrow} \{-1, 1\} \times 1 \to 1$$

and this pushout is preserved by the forgetful functor to $\mathbf{Top}$. It looks like two lines that have been glued together at the origin.

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    $\begingroup$ Did this answer your question? $\endgroup$
    – Todd Trimble
    Jul 23 '13 at 10:37
  • $\begingroup$ I completely forgot the existence of this thread; after your answer I simply thought "Damn, it's false. Nevermind, let's do something else". $\endgroup$
    – fosco
    Oct 30 '13 at 1:09

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