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.
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