Background
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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:



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> [1] Daniel K. Biss, A Generalized Approach to the Fundamental Group, The American Mathematical Monthly, Vol. 107
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You can find this [online](http://math.mit.edu/~gracelyo/18904/BissCoveringSpaceArticle.pdf). This is somehow an introduction to



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> [2] Daniel K. Biss, The topological fundamental group and generalized covering spaces , Topology and its Applications, Vol. 124
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Question
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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](http://www.uni-bonn.de/~habecker/dokumente/tutorium/prodquotient2.ps). Remark however that this is true in the category of compactly generated spaces.



Fundamental group as topological group
**Update**: A bit of a digital paper chase led me, via [David Robert's thesis](http://ncatlab.org/davidroberts/show/HomePage) (note that in the latest version, it is Chapter 5, section 2 that is most relevant), to [this paper](http://arxiv.org/abs/0910.3685) on the arXiv. The last sentence of the abstract is:



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> These hoop earring spaces provide a simple class of counterexamples to the claim that $\pi\_{1}^{top}$ is a functor to the category of topological groups.
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I recommend reading this article.


(**Added later**: In case it's not clear, the author of that paper is Jeremy Brazas who added an answer afterwards, so if you vote for my answer, you should *definitely* vote for his!)




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**Original Answer**: *These were my initial thoughts before I found the references above. These were what made me sufficiently intrigued to do the paper chase and find the above-mentioned thesis and article.*


The proof given in the second paper (by Biss) that is mentioned in the question is short enough that I think it reasonable to copy it out here. I shan't copy out the obvious diagram so need to establish some notation first:


1. $m \colon \pi\_1^{Top}(X,x) \times \pi\_1^{Top}(X,x) \to \pi\_1^{Top}(X,x)$ is the multiplication map in question
2. $p \colon \operatorname{Hom}((S^1,1),(X,x)) \to \pi\_1^{Top}(X,x)$ is the quotient map
3. $\overline{m} \colon \operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x)) \to \operatorname{Hom}((S^1,1),(X,x))$ is the "upstairs" multiplication map. (It's a tilde in the original, but that isn't displaying correctly for me so I daren't use it.)


The proof then proceeds:



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> To show that $m$ is continuous, it suffices to show that $\overline{m}$ is continuous, for then if $U \subset \pi\_1^{Top}(X,x)$ is open, $(p \times p)^{-1} m^{-1}(U) = \overline{m}^{-1}p^{-1}(U)$ is open, but by the definition of a quotient map, $(p \times p)^{-1} m^{-1}(U)$ is open if and only if $m^{-1}(U)$ is.
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There then follows a proof that $\overline{m}$ is continuous, a fact that I trust does not need proving.


Comments on your comments:


1. We don't need the quotient map to be open since we are only ever dealing with preimage sets. It is certainly not always true that if $q \colon X \to Y$ is a quotient that $q(U)$ is open in $Y$ for every open $U$ in $X$. But it is true *by definition* that $q^{-1}(U)$ is open in $X$ **if and only if** $U$ is open in $Y$. This is because the topology on $Y$ is precisely that to make this true. So since we are only dealing with sets of the form $(p \times p)^{-1}(A)$ then the assertion is valid *assuming that $p \times p$ is a quotient map*.
2. Here, I find myself worried. A quick back-of-envelope check seems to show that one can't simply assume that the product of quotients is again a quotient in Top (a counterexample eludes me as I don't have *Counterexamples in Topology* to hand and I'm too used to dealing with "nice" spaces). It *may* be the case that for Hom-spaces then there's some magic that can be done (though such is not mentioned in the paper); but again the best that I can do on the back of an envelope is observe that (modulo some basepoint mess) *by construction* $\operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x))$ quotients to $\pi\_1^{Top}((X,x) \times (X,x))$. But to proceed, one would need to know that $\pi\_1^{Top}$ was a product-preserving functor. This is morally the same as saying that it is representable - which looks good since we have an obvious representing object $S^1$! However, this can't be made into a proper argument since although we have a representing object, we *don't* have an enriched Hom-functor $hTop \times hTop \to Top$ which to evaluate at $S^1$.


So I would look for a counterexample to the product of quotients being a quotient, and see where that leads you. Either you'll find a proper counterexample to the proposition in question, or you'll see why *in this special case*, such a counterexample could not occur.


(Of course, I may well be missing something obvious!)


