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I feel I want to understand it better. I know that for every cover there's a finite subcover but what can you say about it's group properties?

I'm stuck on this homework problem where we were asked: Let G compact, Hausdorff which has the structure of a group. And multiplication $m: G \times G \rightarrow G$ is continuous. Show G is a topological group.

All I need to do is to show the inverse map, $ inv: G\to G $ is continuous. So I have to somehow use multiplication is continuous and the fact it is compact to show inv is continuous.

I tried to reason $inv(x) = x^{-1}$ as $L_{x^{-2}}(x) = x^{-2}*x = x^{-1}$ and try to say something with continuity since $L_{x^{-2}}$ is continuous, guaranteed by multiplication is continuous. But $inv$ becomes too independent on a particular x it operates on... So I don't know.

I think if I just know compact topological groups better I'd be in okay shape.

Thanks!

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Standard homework probably does not belong in MO. What else do you know about "compact Hausdorff" besides the definition? –  Gerald Edgar Nov 14 '10 at 22:44
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The question has been closed. In level, its appropriateness is borderline (and thus it has received a correct answer). However, it has been phrased purely as a homework problem, with no other motivation, and such questions are discouraged on MO. (It would be a perfectly good question on math.stackexchange.com, for instance.) –  Pete L. Clark Nov 14 '10 at 23:40
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closed as off topic by Gerald Edgar, Qiaochu Yuan, Ryan Budney, Andreas Thom, Pete L. Clark Nov 14 '10 at 23:38

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1 Answer

Consider the function $\beta:(x,y)\in G\times G\mapsto (x,xy)\in G\times G$. It is continuous and bijective, as you can easily check. It follows from your hypotheses too that $\beta$ is an homeomorphism. If now we define the functions $\lambda:x\in G\mapsto (x,1)\in G\times G$ and $\pi:(x,y)\in G\times G\mapsto y\in G$, which are obviously continuous, then the composition $\iota=\pi\circ\beta^{-1}\circ\lambda:G\to G$ is also continuous.

Magically, $\iota$ is the inversion map.

(I learnt this argument from reading the extraordinary proof by Peter Schauenburg that a bialgebra $H$ over a field $k$ for which there exists an $H$-Galois extension of $k$ is in fact a Hopf algebra. In a way, this is the non-non-commutative version :) )

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Dear Mariano, I suppose there are easy counterexamples if $G$ is not compact, but I don't see any. Can you please help ? –  Georges Elencwajg Nov 15 '10 at 14:00
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Dear Georges, if the underlying space is locally compact and Hausdorff, we also have a topological group provided the product is continuous (Ellis, Robert. Locally compact transformation groups. Duke Math. J. 24 (1957), 119--125. MR0088674) and if the space is metric, complete and separable and the left and right translation functions are continuous (which is weaker that having the product be continuous) then we also have a topological group (Montgomery, Deane. Continuity in topological groups. Bull. Amer. Math. Soc. 42 (1936), no. 12, 879--882. MR1563458) Examples are bound to be ugly! –  Mariano Suárez-Alvarez Nov 15 '10 at 14:54
    
Thank you for your (as usual) erudite answer, Mariano. –  Georges Elencwajg Nov 18 '10 at 7:42
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