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



The Tychonov cube $X^X$ of a compact space $X$ is a compact semigroup with the composition operation
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).


