Making the subgroup of units of a topological ring a topological group 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.
 A: 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
