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Let $H$ denote a group that is also a Hausdorff topological space. Show that $H$ is a topological group if and only if the map $h: H \times H \rightarrow H, h:=(x,y)\rightarrow xy^{−1}$ is continuous.

The following proof does not use the Hausdorff property.

Suppose $h$ is continuous. 

$f:H \rightarrow H \times H, f(x):=(e, x)$ is continuous (Munkres theorem 19.6). $(h\circ f)(x)=x^{−1}$ is continuous. 

$j:H\times H \rightarrow H \times H, j(x, y):=(x, y^{−1})$. $\pi _1 (x, y)=x$ is continuous. $(h\circ f \circ \pi _2)(x, y)=y^{−1}$ is continuous. By Munkres theorem 19.6, $j$ is continuous. $(h\circ j)(x, y)=xy$ is continuous.

So $H$ is a topological group.

Suppose $H$ is a topological group. $k(x, y):=xy$ and $l(x):=x^{−1}$. 

$\pi _1 (x, y)=x$ is continuous. $(l \circ \pi _2) (x, y)=x^{−1}$ is continuous. By Munkres theorem 19.6, $j$ is continuous. $j \circ k=h$ is continuous.

Please let me know whether Hausdorff property is required. (I'm aware that a topological group is Hausdorff).

Thanks.

Edit: Thanks to Aryaman Maithani's comment, some textbooks (for eg Munkres) require Hausdorff-ness in the definition of topological group. Hausdorff-ness is required probably only in that case.

Vinay Deshpande
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    Looks like Munkres requires $T_1$ as part of the definition for a topological group. If you drop the Hausdorf-ness, how do you get that $H$ is $T_1$? – Aryaman Maithani Feb 17 '21 at 08:37
  • The Hausdorff requirement is just to satisfy that extra separation axioms requirement; note that if $G$ is a topolgoival group in the "continuous operations" sense, $T_0$, $T_1$, Hausdorff, $T_3$ and $T_{3\frac12}$ are all equivalent anyway. Demand one, get the others for free.. – Henno Brandsma Feb 17 '21 at 09:11

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Indeed the proof does not require any Hausdorffness, we only need that certain other natural product maps are continuous, and this holds regardless of separation axioms.

I would do it this way: suppose $h$ is continuous, then indeed $f:x \to (e,x)$ from $H$ into $H \times H$ is continuous (the composition with the two projections are a constant map (always continuous) resp the identity on $H$ (ditto). The inversion map $I:x \to x^{-1}: H \to H$ then is continuous as $h \circ f$, a composition of continuous maps. The product map $P: (x,y) \to xy$ is then also continuous as $h \circ (1_H \times I)$ etc.

And if $P$ and $I$ are continuous so is $h$ as the composition $P \circ (1_H \times I)$, so both implications hold.

Henno Brandsma
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