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According to Wikipedia, a topological group $G$ is a group and a topological space such that $$ (x,y) \mapsto xy$$ and $$ x \mapsto x^{-1}$$

are continuous. The second requirement follows from the first one, no? (by taking $y=0, x=-x$ in the first requirement)

So we can drop it in the definition, right?

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No we can't drop it in general, see e.g. the discussion here. – t.b. Jul 19 '12 at 18:37
@t.b. Right thanks, see my comment to the answer. – Rudy the Reindeer Jul 19 '12 at 18:38
@bananalyst what I wrote, and even what I wanted to write, made no sense ^^ – Olivier Bégassat Jul 19 '12 at 18:40
@OlivierBégassat Ok, I will delete obsolete comment : ) – Rudy the Reindeer Jul 19 '12 at 18:41
up vote 1 down vote accepted

the actual requirement is that $f: G \times G \to G, \ f(x,y) = xy$ is continuous in the product topology, so 'taking $x = -x$' doesn't make any sense.

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But in the definition of a topological vector space $x \mapsto x^{-1}$ is not part of the definition but surely every TVS is also a topological group. – Rudy the Reindeer Jul 19 '12 at 18:37
while writing it became clear: we get the second requirement from $\alpha = -1$ for scalar multiplication. – Rudy the Reindeer Jul 19 '12 at 18:37
The definition of topological vector space requires that both $+\colon V\times V \to V$, $+(a,b) = a+b$ and $\cdot\colon K\times V\to V$, $\cdot(\alpha,a) = \alpha a$ be continuous. So you are still requiring that two functions be continuous; it's not enough for the addition to be continuous, you need something else. The requirement for a topological vector space is stronger than the requirement for a topological group (since as you note, you get continuity of $x\mapsto -x$ from restriction of the scalar multiplication map). – Arturo Magidin Jul 19 '12 at 18:51

"Continuity of $f\colon G\times G\to G$, $f(x,y)=xy$" means that the function is continuous relative to the topology of $G$ and the induced product topology of $G\times G$.

It is true that when you have a continuous function $g\colon X\times Y\to Z$, where $X$ and $Y$ are topological spaces and the topology on $X\times Y$ is the product topology, then for each fixed $x\in X$ the induced map $g_x\colon Y\to Z$ given by $g_x(y) = g(x,y)$, and for each fixed $y\in Y$, the induced map $g_y\colon X\to Z$ given by $g_y(x) = g(x,y)$, are continuous.

However, you cannot obtain continuity of $h(x)=x^{-1}$ here in that way. You need the map to take $x$ as an input, and produces $x^{-1}$ as an output. Your suggestion is to actually take the function $m_{e}(f(x))$, where $m$ is the multiplication map, $f$ is the inversion map, and $e$ is the identity; in order to deduce that this map is continuous from the continuity of $m$, you essentially have to show that $f$ itself is continuous... which is what you are trying to establish in the first place.

So continuity of $f$ does not immediately follow as you suggest.

For an example to show that you cannot deduce continuity of $x\longmapsto x^{-1}$ merely from the continuity of $(x,y)\longmapsto xy$, see for example this answer by tomasz: take $G=\mathbb{R}$ under addition, with the Sorgenfrey topology: a basis for the open sets consist of the intervals of the form $[a,b)$. Then multiplication is continuous, but the inverse map is not, since the pullback of $[a,b)$ is $(-b,a]$, which is not open in the topology.

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But we can get continuity of $x^{-1}$ from continuity of scalar multiplication in TVSs! – Rudy the Reindeer Jul 19 '12 at 18:46
@bananalyst: But this is not a topological vector space, so how is your comment relevant? Scalar multiplication is not continuous under the Sorgenfrey topology. – Arturo Magidin Jul 19 '12 at 18:47
Sorry. The reason why I thought about this is because I noticed that the second requirement is missing from definition of TVS. So I started to wonder whether it couldn't be dropped before I realised that it is contained in continuity of scalar multiplication. – Rudy the Reindeer Jul 19 '12 at 18:51

There is even a term for a group endowed with a topology such that multiplication is continuous (but inversion need not be): a paratopological group.

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Cool, thanks for pointing this out! – Rudy the Reindeer Jul 19 '12 at 19:11
And as is said on the Wikipedia page, compact paratopological groups are automatically topological groups. – polmath Oct 3 '14 at 17:57

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