Can continuity of inverse be omitted from the definition of topological group?

Question

According to Wikipedia, a topological group $G$ is a group and a topological space such that $$ (x,y) \mapsto xy$$ and $$ x \mapsto x^{-1}$$

are continuous. The second requirement follows from the first one, no? (by taking $y=0, x=-x$ in the first requirement)

So we can drop it in the definition, right?

Answer 1

the actual requirement is that $f: G \times G \to G, \ f(x,y) = xy$ is continuous in the product topology, so 'taking $x = -x$' doesn't make any sense.

Answer 2

"Continuity of $f\colon G\times G\to G$, $f(x,y)=xy$" means that the function is continuous relative to the topology of $G$ and the induced product topology of $G\times G$.

It is true that when you have a continuous function $g\colon X\times Y\to Z$, where $X$ and $Y$ are topological spaces and the topology on $X\times Y$ is the product topology, then for each fixed $x\in X$ the induced map $g_x\colon Y\to Z$ given by $g_x(y) = g(x,y)$, and for each fixed $y\in Y$, the induced map $g_y\colon X\to Z$ given by $g_y(x) = g(x,y)$, are continuous.

However, you cannot obtain continuity of $h(x)=x^{-1}$ here in that way. You need the map to take $x$ as an input, and produces $x^{-1}$ as an output. Your suggestion is to actually take the function $m_{e}(f(x))$, where $m$ is the multiplication map, $f$ is the inversion map, and $e$ is the identity; in order to deduce that this map is continuous from the continuity of $m$, you essentially have to show that $f$ itself is continuous... which is what you are trying to establish in the first place.

So continuity of $f$ does not immediately follow as you suggest.

For an example to show that you cannot deduce continuity of $x\longmapsto x^{-1}$ merely from the continuity of $(x,y)\longmapsto xy$, see for example this answer by tomasz: take $G=\mathbb{R}$ under addition, with the Sorgenfrey topology: a basis for the open sets consist of the intervals of the form $[a,b)$. Then multiplication is continuous, but the inverse map is not, since the pullback of $[a,b)$ is $(-b,a]$, which is not open in the topology.

Answer 3

There is even a term for a group endowed with a topology such that multiplication is continuous (but inversion need not be): a paratopological group.