Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a topological group acts on the topological space $X$, for an elememt $g\in G$, let's define the map $f:X\rightarrow X, f(x)=g*x$.

I am trying to find if $f$ is continuous?

my best thanks

share|cite|improve this question
Without further condition on the group action, $f_g$ is not necessarily continuous. Usually, with topological groups, you specify that the group action: $\phi:G\times X\rightarrow X$ is continuous. Then $f_g(x)=\phi(g,x)$ is continous. – Thomas Andrews Oct 19 '12 at 0:09
yes, actually $\phi:G\times X \rightarrow X$ is continuous, but how can we prove that $f_{g}:X\rightarrow X$ defined above is continuous – kiranovalobas Oct 19 '12 at 0:16
Because $h_g:X\to G\times X$ defined as $h_g(x)=(g,x)$ is continuous. And $f_g=\phi \circ h_g$ – Thomas Andrews Oct 19 '12 at 0:22

When we say $G$ acts on some object $X$, we usually mean that each group element $g$ is assigned a map $f_g:X\to X$. The nature of this map depends on the context of the object. We should ask: To which category does $X$ belong? The category determines what the objects and maps are, and how maps are composed with one another. If $X$ is a set, the maps are ordinary functions. If $X$, however, is a topological space, the maps we are interested in are usually the continuous maps. So when $G$ acts on a topological space $X$, it is usually assumed that each $f_g$ is continuous.

This is automatic if the group action is continuous, meaning that the map $G\times X\xrightarrow \mu X, (g,x)\mapsto f_g(x)$, is continuous. In that case, for a fixed $g\in G$, the function $f_g$ is the composite $\mu\circ\text{in}_g$, where $\text{in}_g(x)=(g,x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.