Let \(H\) denote a group that is also a topological space satisfying the \(T_1\) axiom. Show that \(H\) is a topological group if and only if the map of \(H \times H\) into \(H\) sending \(x \times y\) into \(x \cdot y^{-1}\) is continuous.
Solution
Let \(\mu(x,y)=xy\), \(\iota(x)=x^{-1}\), and \(\phi(x,y)=xy^{-1}\). A \(T_1\) group \(H\) is a topological group when \(\mu\colon H\times H\to H\) and \(\iota\colon H\to H\) are continuous. A map into \(H\times H\) is continuous if and only if its two coordinates are.
1. If \(\mu\) and \(\iota\) are continuous, then \(\phi\) is continuous.
\(\phi=\mu\circ(\operatorname{id}_H\times\iota)\), and \(\operatorname{id}_H\times\iota\colon(x,y)\mapsto(x,y^{-1})\) is continuous because its coordinates \(\pi_1\) and \(\iota\circ\pi_2\) are.
2. If \(\phi\) is continuous, then \(\iota\) and \(\mu\) are continuous.
The map \(j(x)=(e,x)\) is continuous, with a constant and an identity coordinate, and \(\iota=\phi\circ j\) since \(\phi(e,x)=x^{-1}\). Then \(\operatorname{id}_H\times\iota\) is continuous, and \(\mu=\phi\circ(\operatorname{id}_H\times\iota)\) since \(\phi(x,y^{-1})=xy\).
3. Q.E.D.
Steps 1 and 2 give the two implications.