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I'm a beginner in the theory of topological algebra (I'm reading something about it in Robert's "A course in $p$- adic analysis"). At page 24, the author states that if $A$ is a topological ring, the subgroup $A^{\times}$ is not in general a topological group.

In order to overcome this difficulty, the author considers the embedding $$x\mapsto (x,x^{-1})\colon A^{\times} \to A\times A$$ and defines the initial topology on $A^{\times}$, i.e. the coarsest topology making the above function continuous.

At this point, he says that $A^{\times}$ is a topological group because the continuity of the inverse map, induced by the symmetry $(x,y)\mapsto (y,x)$ of $A\times A$, is obvious.

Well, it is not so clear to me why $A^{\times}$ is a topological group.

Attempt: if $f$ denotes the above embedding and $U(x)$ an open nhbd of $x\in A$, an open nhbd in $A^{\times}$ in the jnitial topology has the form $$f^{-1}(U(x)\times U(x^{-1}))=\{z\in A^{\times}\mid (z,z^{-1})\in U(x)\times U(x^{-1})\}.$$

Hence $$f^{-1}(U(xy)\times U(y^{-1}x^{-1}))=\{(z,w)\in (A^{\times})^2\mid (zw,w^{-1}z^{-1})\in U(xy)\times U(y^{-1}x^{-1})\}.$$ (NOTE: Maybe it is convenient to consider non-empty symmetric open sets $V(x)=U(x)^{-1}\cap U(x^{-1})$, if it does make sense, because in this case $V(x)^{-1}=V(x^{-1})$.) Now, I don't know how to write this preimage as union of product of preimages of basic open sets so that continuity is proved.

Thank you in advance for your help.

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The topology you define on have can also be defined by a universal property that will allow you to prove all you want pretty magically.

Indeed, its definition shows that a map $X\to A^\times$ is continuous if and only if the composite $X\to A\times A$ is continuous. Try to prove that if it isn't clear for you.

With this in mind, we have two maps whose continuity we want to prove : multiplication and inversion, both have codomain $A^\times$ so that's great. But now look at inversion for instance : $x\mapsto x^{-1}$. Then the composite $A^\times \to A\times A$ is $x\mapsto (x^{-1},x)$, which is $sym\circ embedding$, both of which are continuous.

What about multiplication then ? Well compose $A^\times\times A^\times \to A^\times \to A\times A$. This is exactly $(x,y) \mapsto xy \mapsto (xy, y^{-1}x^{-1})$.

But magically, this is only the composition $A^\times \times A^\times \to A\times A\times A\times A\to A\times A\times A\times A \to A\times A$ where the middle map is $(a,b,c,d) \mapsto (a,c,d,b)$ and the last map is $(a,b,c,d)\mapsto (ab, cd)$ so everything is continuous. Therefore the composition is continuous as well, and we are done

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  • $\begingroup$ I try to prove $(\Leftarrow)$ of the Universal property: if $U$ is open in $A^2$, $f$ is the embedding and $g\colon X\to A^{\times}$, then $f^{-1}(U), (f\cdot g)^{-1}(U)=g^{-1}[f^{-1}(U)]$ are opne and the conclusion follows. Universal properties seem to be very useful... do you know good references for studying them? $\endgroup$ – LBJFS Mar 21 at 14:36
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    $\begingroup$ @LBJFS : universal properties are indeed very powerful when you're used to manipulating them and can recognize them from a distance ! :) I don't know any reference specifically about them, to learn about them in general you'll have to look for category theory documents. You may start with the wikipedia page : en.wikipedia.org/wiki/Universal_property which already contains plenty of information $\endgroup$ – Max Mar 21 at 14:53
  • $\begingroup$ Great! Thank you :) $\endgroup$ – LBJFS Mar 21 at 14:56

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