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A metrizable group is a metric space $(G,d)$ with a binary operation $\cdot$ such $(G,(\cdot))$ is a group and maps $(\cdot):G\times G\to G$ and $f:G\to G$ given by $(\cdot)(x,y)=xy$ and $f(x)=x^{-1}$ are continuous respect $d$.

Why requires definition that $f$ be continuous? Is it possible to have continuity of $(\cdot)$ without continuity of $f$?

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    I don't understand your last sentence. Of course, $y^{-1}$ exists; the question is whether the inversion map is continuous.2012-05-31
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    But.. inverses are related with the binary operation, I don't know why need an aditional condition.2012-05-31
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    (i) Actually, formally a group is defined by *three* operations: a binary operation (the product), a *unary* operation (inversion), and a *nullary* operation (the identity element). (ii) Are you asking whether the continuity of inversion follows from continuity of the product?2012-05-31
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    Yes, they are "related", but are they "related" enough that the continuity of the binary operation implies the continuity of the inversion map? I think the answer is no although I don't have a counterexample at hand.2012-05-31
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    @ArturoMagidin of course, identity continuity follow from product ontinuity, do you say that existence of inverses and null elements induces a new opration?. I think that it is right but I didn't see with this point of view. Thanks! Why need continuity of inverse, not enought with product (and identity induced by product)?2012-05-31
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    @GastónBurrull: First, no, they don't "induce new operations". They are **part of the definition** of a group when you describe a group as an algebra. Second: the point is that if you really want to define a topological group, you need to consider *all* operations as functions; the inverse function cannot be expressed that way in terms of the product function alone (you need existential quantifiers to describe it). The nullary function is continuous because it is constant, so you don't need to put conditions on it. So your question is *really* "does continuity of $\cdot$ imply that of $f$?"2012-05-31
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    @ArturoMagidin Yes, you're right. The other question have not sense. I dont know a counterexample when we have continuity of $\cdot$ and $f$ is discontinuous in some point. I saw that continuity of product implies continuity of identity, but is because definition of group we dont need continuity of binary operation to conclude continuity of identity as you said.2012-05-31
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    @GastónBurrull: Perhaps you could edit the question and clarify it, then, so people don't have to dig through the comments to find the explanations?2012-05-31
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    @ArturoMagidin Thanks for your help. Your answers are so clear. Are you a mathematical professor?2012-05-31

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I made a mistake here before. Sorry!

Continuity of multiplication is not sufficient for general topological groups.

Consider the standard additive group of reals with Sorgenfrey topology. Inverse is clearly not continuous, since $-[a,b)=(-b,-a]$.

Multiplication (or, rather, addition) is continuous, since, intuitively, the open sets in the product topology are the sets which do not have "upper" and "right" edge (but may possibly have "left" and "lower" edge, or some parts of those). More precisely, for any $x+y=c\in [a,b)$, then let $0<\varepsilon<(b-c)/2$. Then for any $(x',y')\in [x,x+\varepsilon)\times [y,y+\varepsilon)$, $a\leq x+y\leq x'+y' , so $(+)^{-1}[a,b)$ is open.

EDIT: I found an example in Topological Groups and Related Structures by Archangel'skii & Tkachenko, example 3.5.6 on pages 175-176. It is the group of homeomorphisms of the space $\lbrace n,0,1/n\mid n\in \mathbf N\rbrace$ with the natural topology, the compact-open topology.

It is metrizable, but not metrizable by a left- or right- invariant metric, and the inverse is not continuous.

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    I need more time to full understand your answer. I'll acept when I'll totally sure. Thanks!2012-05-31
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    I think things are OK for completely metrizable spaces. Forget where I saw it.2012-05-31
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    To follow up on André's comment, continuity of inversion follows from continuity of multiplication in quite some generality by a Baire category argument, e.g. for locally compact Hausdorff or completely metrizable spaces with a group structure such that multiplication is (separately) continuous: Pfister, *[Continuity of the inverse](http://dx.doi.org/10.1090/S0002-9939-1985-0801345-5)*, Proc. Amer. Math. Soc. **95** (1985), 312-314 contains the best results in this direction I'm aware of.2012-05-31
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    @t.b. Thanks a lot!!. Must you mean that be metrizable is not sufficient condition?.2012-05-31
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    @Gastón: No, metrizability is not a sufficient condition for continuity of the multiplication to imply continuity of the inversion. tomasz made an edit to the answer giving a reference to a counterexample.2012-06-01
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Take $G=\mathbb R$ with addition and put on it the topology which has the set $\{(-\infty,a):a\in\mathbb R\}$ as a basis. Then addition is continuous but inversion is not.

You want it metrizable, though...

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    Probably your example is the easier when inversion continuity not follow from other in topological spaces. Sometimes I think that metrizable spaces are not enough studied like other topological spaces.2012-05-31