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.

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

• 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? – LBJFS Mar 21 at 14:36
• @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 – Max Mar 21 at 14:53
• Great! Thank you :) – LBJFS Mar 21 at 14:56