Equivalence to being a topological group

Question

Just some notation I am using:

A topological group $G$ is a group with a topology such that

$o : G^2 \to G : (x,y) \mapsto xy$

and

$inv : G \to G : x \mapsto x^{-1}$

are continuous in the associated topology.

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From Munkres:

Show that $G$ is a topological group if the map of $G \times G$ into $G$ sending $x \times y$ into $x^{-1} \cdot y$ is continuous, and conversely.

Can I be the first to say I find Munkres to be incredibly awkward in the wording of his questions?

I think he wants me to prove the following statement:

$o$ and $inv$ are continuous $\Longleftrightarrow$ $\bar o : G^2 \to G : (x,y) \mapsto x^{-1}y$ is continuous.

I see the proof for the $\Longrightarrow$ direction, however I don't see how to prove the other way, which leads me to the idea that maybe I misunderstand what I need to prove.

Any hints would be appreciated!

Answer 1

Hint:
(1) First, show that, or note that $x\mapsto (x,e)$ is continuous.
(2) Show that $x\mapsto x^{-1}$ is continuous, by showing that it is the composition of two continuous map.
(3) Show that $(x,y)\mapsto xy$ is continuous, by the same method of (2).

Answer 2

Hint:

Consider the maps \begin{alignat*}{2} G&\longrightarrow G\times G&&\longrightarrow G\\ x&\longmapsto(x,1)&&\longmapsto x^{-1}\cdot 1 \end{alignat*} then \begin{alignat*}{2} G\times G&\longrightarrow G\times G&&\longrightarrow G\\ (x,y)&\longmapsto(x^{-1},y)&&\longmapsto (x^{-1})^{-1}y=xy\\ \end{alignat*}