# 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.

## 1 Answer

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