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Let $X$ and $Y$ be topological spaces. Define the compact open topology on the set $\mathrm{M}(X,Y)$ of contiunous maps from $X$ to $Y$ via the subbase $[K,O]$ of all maps $f:X\rightarrow Y$ s.t. $f(K)\subset O$, where $K$ is any compact subset of $X$, and $O$ is any open subset of $Y$. So a basis of open sets is given by the following subsets: $[K_1,\dots,K_n,O_1,\dots,O_n]=[K_1,O_1 ]\cap\dots\cap [K_n,O_n]$, the collection of continuous maps $f:X\rightarrow Y$ that send each $K_i$ into $O_i$ for some specified collcetion of compact $K_i$'s and open $O_i$'s.

This topology has some nice properties: the exponential law holds under some hypotheses on the spaces $X$ and $Y$, and is certainly true if all spaces involved are locally compact Hausdorff spaces, as will be the case from now on.

My question is as follows: if $X$ is a locally compact Hausdorff space (or even a topological manifold), the compact open topology induces a topology on the set of homeomorphisms of $X$, which is a group. Does this topology turn $\mathrm{Homeo}(X)$ into a topological group? I can show that the product (composition) is continuous, but is the inverse too? $(f\rightarrow f^{-1})$

I was able to prove continuity for compact spaces, where it is very easy to establish. I also managed to prove it for $X=\mathbb{R}$ because all homeomorphisms of $\mathbb{R}$ are monotone, but that's everything so far.

I tried looking it up in several textbooks on topology and algebraic topology where the C.O. topology is usually discussed, but couldn't find a discussion on this topic anywhere.

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1+, interesting. Why does it work for compact $X$? –  Martin Brandenburg Mar 16 '11 at 23:07
@Martin: suppose $X$ is compact and $f$ is homeomorphic. $f \in [K, O]$ iff $f^{-1} \in [X\O, X\K]$. $X\O$ is compact, and all compact subsets of $X$ can be expressed in this way. I haven't written down the argument, but it feels like it should follow trivially from this observation and the definition of the topology. –  Tom Ellis Mar 16 '11 at 23:27
I did this a couple of years back, but it's really easy and involves \emph{no} topology at all, which came as a real surprise. Let's call $\mathrm{inv}:\mathrm{Homeo}(X)\rightarrow\mathrm{Homeo}(X)$ the inversion of homeomorphisms. We want to show that this defines a continuous map and even a homeomorphism whenever $X$ is compact. $f$ belongs to $\mathrm{inv}([K,O])$ iff $f^{-1}$ sends $K$ into $O$ iff $K$ is a subset of $f(O)$ iff $X\setminus f(O)=f(X\setminus O)$ is a subset of $X\setminus K$ iff $f$ belongs to $[X\setminus O, X\setminus K]$ inv is an open bijection and is its onw inverse –  Olivier Bégassat Mar 16 '11 at 23:30

4 Answers 4

up vote 13 down vote accepted

The following article gives you a lot of information on the question you are asking:

On Homeomorphism Groups and the Compact-Open Topology, Jan J. Dijkstra



The answer is in general "no".

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thank you for the link, I'm very excited to read this short article, I love the idea to replace the locally compact hausdorff space $X$ by its one point compactification, where the theorem works. Thanks again! –  Olivier Bégassat Mar 16 '11 at 23:35
See also Bourbaki, Topologie générale, chapitre X, §3. Especially no. 5, Topologie sur les groupes d'homéomorphismes. (Proposition 11, page 30 says it works for a compact space, and the following pages explain why it fails in general, using the one-point compactification.) On the same page, Bourbaki also explains that for a uniform space $X$, a subgroup of $\mathit{Homeo}(X)$ consisting of equicontinuous homeomorphisms is a topological group for the topology of pointwise convergence. –  ACL Mar 17 '11 at 9:28
By the way, Bourbaki's counterexample (page X.51, exercise 17) is the one given by Robert Israel below. –  ACL Mar 17 '11 at 9:30

For a simple counterexample, let X be the subspace of R consisting of 0 and exp(n) for all integers n, and consider the homeomorphisms $f_n$ of X defined by $f_n(0) = 0$, $f_n(\exp(k)) = \exp(k-1)$ for $k \le -n$ or $k > n$, $\exp(k)$ for $-n < k < n$, and $\exp(-n)$ for $k = n$. Then $f_n$ converges to the identity map in the compact-open topology, but $f_n^{-1}$ does not: $f_n^{-1}$ is not in the neighbourhood of the identity map given by $K = X \cap [0, 1]$ and $U = X \cap (-1, 2)$ for any $n \ge 1$, because $\exp(-n) \in K$ and $f_n^{-1}(\exp(-n)) = \exp(n) \notin U$.

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R. Arens, Topologies for homeomorphism groups, Amer. J. Math. 68 (1946) 593–610.

If $X$ is locally compact and \emph{locally connected} (!), then $Homeo(X)$ is a topological group.

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I would also recommend the very nice paper: On a Class of Transformation Groups Andrew M. Gleason and Richard S. Palais American Journal of Mathematics Vol. 79, No. 3 (Jul., 1957), pp. 631-648 –  Claudio Gorodski Jun 4 '11 at 0:10

I don't know if this helps, but you could consider the set $[K,O]'$ of all maps $f\in Homeo(X)$ such that $f(K)\subset O$ and $f^{-1}(K)\subset O$. With the sets $[K,O]'$ as a subbase of neighborhoods of the identity, $Homeo(X)$ becomes a topological group, if $X$ is a locally compact $T_3$-space. This topology is called the Braconnier topology. It coincides with the compact open topology whenever $X$ is compact, or locally connected. In general it is stronger than the compact open topology.

Unfortunately I don't know any references for this facts, except the lecture notes of the 1982/83 lectures of Prof. Holdgruen on harmonic analysis (in a shelf in the library of the mathematics institute in Goettingen), because I learned this in my undergrad courses and not from books.

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