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I have a question regarding continuity in topological groups. From Wikipedia:

A topological group $G$ is a topological space that is also a group such that the group operation:

\begin{align*} \mu: G \times G &\rightarrow G\\ (x,y) & \mapsto xy \end{align*}

and inversion map:

\begin{align*} \iota: G &\rightarrow G\\ x & \mapsto x^{-1} \end{align*}

are continuous. The product map is continuous iff for any $x,y \in G$ and any neighborhood $W$ of $xy$ in $G$, there exist neighborhoods $U$ of $x$ and $V$ of $y$ in $G$ such that $U\cdot V \subseteq W$.

Now what bothers me is that the real definition of continuity for this operation should be:

The product map is continuous iff for any $x,y \in G$ and any neighborhood $W$ of $xy$ in $G$, there exists a neighborhood $U\times V$ of $(x,y)$ in $G \times G$ such that $U\times V \subseteq \mu^{-1}(W)$.

Why is not this the definition? Is there some "trivial" way to go from one to the other?

user123456
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1 Answers1

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Yes. Note that $U \cdot V \subseteq W$ is the same as saying that $\mu(U \times V) \subseteq W$. This is indeed equivalent to $U \times V \subseteq \mu^{-1}(W)$.

Indeed, $f(X) \subseteq Y$ and $X \subseteq f^{-1}(Y)$ are equivalent statements in general for any function $f$, just by unpacking the definition of an inverse function.

Daniel McLaury
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