2
$\begingroup$

Reading a book about Ramsey theory this is the first example of a compact (semitopological) semigroup, which is a nonempty semigroup S with compact Hausdorff topology for which $x \mapsto x*s$ is a continuous map for all $s$ in $S$

If $X$ is a compact Hausdorff space, then the Tychonov cube $X^X$ is a compact (semitopological) semigroup with the composition from the left operation, $f \mapsto f \circ g$. Note that in general the operation of composition from the right $f \mapsto g \circ f$ is not necessarily continuous on $X^{X}$ unless $g: X \rightarrow X$ is continuous.

I've been trying to prove that the composition from the left is continuous but i still cant, a hint would help me a lot

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ If $A, B$ are sets, $X$ is a topological space, and $\alpha\colon A\to B$ is any function then the induced map $X^B\to X^A$, defined by $f\mapsto f\circ \alpha$ is continuous. This follows from the universal property of product. Taking $A=B=X$ you get the result that you want. $\endgroup$
    – Gregory Arone
    Commented Jan 15, 2021 at 7:00
  • 1
    $\begingroup$ By definition a map $Y\to X^X$ is continuous iff for any $x\in X$ the composition with the evaluation at $x$ (projection on the $x$-coordinate of the product) is continuous. Thus the continuity of $X^X\ni f\mapsto f\circ g\in X^X$ is obvious, since $X^X\ni f\mapsto (f\circ g)(x)\in X$ is just $f\mapsto f(g(x))$, the evaluation at $g(x)$. $\endgroup$
    – Pietro Majer
    Commented Jan 15, 2021 at 23:45

1 Answer 1

Reset to default
3
$\begingroup$

To check that $f \to f \circ g$ is continuous in $f$ as a map $X^X \to X^X$ for a fixed $g \in X^X$: take a net $f_i \to f$ ($i \in I$, some directed set) in $X^X$ converging to $f \in X^X$. This means exactly that $$\forall x \in X: f_i(x) \to f(x)\tag{1}$$ in $X$. So in particular for any $x \in X$, $(f_i \circ g)(x) = f_i(g(x)) \to f(g(x)) = (f \circ g)(x)$, applying $(1)$ to $g(x)$ as the evaluation point. This means exactly that $(f_i \circ g) \to (f \circ g)$ in $X^X$ by the characterisation of the product topology by pointwise convergence (of nets). So indeed $f \to f \circ g$ is continuous on $X^X$.

Alternatively, in terms of subbases: a standard subbasic element of $X^X$ is of the form $\pi_x^{-1}[O]$ with $O$ open in $X$ (with $\pi_x:X^X \to X, f \to f(x)$ a standard projection for $x \in X$) and then, denoting the composition on the right by $g$ as the map $R_g$, say, we see that

$$R_g^{-1}[\pi_x^{-1}[O]] = \{f \in X^X\mid R_g(f) \in \pi_x^{-1}[O]\}= \{f \in X^X\mid f(g(x))\in O\}=\pi_{g(x)}^{-1}[O]$$

which is also subbasic open, and so $R_g$ is continuous that way too. Composing by $g$ on the left we have

$$L_g^{-1}[\pi_x^{-1}[O]] = \{f \in X^X\mid L_g(f) \in \pi_x^{-1}[O]\} = \{f \in X^X\mid g(f(x)) \in O\} = \\= \{f \in X^X\mid f(x) \in g^{-1}[O]\} = \pi_x^{-1}[g^{-1}[O]]$$

which is then also subbasic open, assuming indeed that $g$ is continuous. The nets proof can be similarly adapted for that second case (left as an easy exercise).

Improve this answer
$\endgroup$
1
  • $\begingroup$ Note that all this works for any space $X$ (compactness is not needed at all, but the author wanted a compact example for other reasons probably). Of course $X$ has to be Hausdorff if we want $X^X$ to be. $\endgroup$
    – Henno Brandsma
    Commented Jan 15, 2021 at 14:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .