## Background

Let $(X,x)$ be a pointed topological space. Then the fundamental group $\pi_1(X,x)$ becomes a topological space: Endow the set of maps $S^1 \to X$ with the compact-open topology, endow the subset of maps mapping $1 \to x$ with the subspace topology, and finally use the quotient topology on $\pi_1(X,x)$. This topology is relevant in some situations. A very interesting paper dealing with this topology is:

[1] Daniel K. Biss, A Generalized Approach to the Fundamental Group, The American Mathematical Monthly, Vol. 107

You can find this online. This is somehow an introduction to

[2] Daniel K. Biss, The topological fundamental group and generalized covering spaces , Topology and its Applications, Vol. 124

## Question

How can we prove that $\pi_1(X,x)$ is a topological group? Clearly the inversion map $\pi_1(X,x) \to \pi_1(X,x)$ is continuous, since $S^1 \to S^1, z \mapsto \overline{z}$ is continuous and induces this map. But I don't know how to attack the continuity of the multiplication. It's not hard to see that the multiplication on $map((S^1,1),(X,x))$ is continuous, since it is induced by a fold map $S^1 \to S^1 + S^1$. In order to carry this over to $\pi_1(X,x)$, there are at least two problems which I encounter:

- The quotient map $map((S^1,1),(X,x)) \to \pi_1(X,x)$ may be not open.
- The product of the quotient maps $map((S^1,1),(X,x))^2 \to \pi_1(X,x)^2$ may be not a quotient map.

In [1] it is claimed that $\pi_1(X,x)$ is always a topological group, and this should be proven in [2], but I have no acecss to [2].

An example that products of quotient maps don't have to be quotient maps can be found here. Remark however that this is true in the category of compactly generated spaces.

topology of the quotient uniformityon $\pi_1(X)$ makes into a topological group. This is shown by the argument in Andrew Stacey's answer below, using that the product of two quotient maps is a quotient map in the category of uniform spaces and uniformly continuous maps. See Isbell's "Uniform spaces" (1964), Exercise III.8(c). See also my answer at mathoverflow.net/questions/54391 $\endgroup$ – Sergey Melikhov Feb 6 '11 at 4:09