Let $X$ and $Y$ be topological spaces. Define the compact open topology on the set $\mathrm{M}(X,Y)$ of continuous 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 collection 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.



Compact open topology on $\mathrm{Homeo}(X)$
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


<http://www.cs.vu.nl/~dijkstra/research/papers/2005compactopen.pdf>


<http://www.jstor.org/pss/30037630>


The answer is in general "no".


