Show that if $X$ is discrete, then $\phi$ is continuous.

Question

Let $S(X,X)$ of all mappings of a set $X$ to itself, taken with the topology of pointwise convergence. Define $\phi: S(X,X) \times S(X,X) \to S(X,X)$ with $\phi((f,g))=f \circ g$.

Show that if $X$ is discrete, then $\phi$ is continuous.

Many thanks!

Answer 1

To show the continuity of $\phi$ in a point $(f,g) \in S(X,X)\times S(X,X)$, we must show that for every neighbourhood $N$ in a subbasis of the neighbourhoods of $\phi(f,g)$, the preimage $\phi^{-1}(N)$ is a neighbourhood of $(f,g)$. A subbasis of the neighbourhoods of $f\circ g$ is given by sets of the form $\{h : h(x) \in V\}$, where $x$ ranges over $X$, and $V$ ranges over a subbasis of neighbourhoods of $f(g(x))$. Since $X$ is discrete, that simplifies to $\{h : h(x) = f(g(x))\}$. Fix an arbitrary $x \in X$. Now, a neighbourhood of $g$ is $G_x = \{ g' \in S(X,X) : g'(x) = g(x)\}$, and a neighbourhood of $f$ is $F_{g(x)} = \{ f' \in S(X,X) : f'(g(x)) = f(g(x))\}$, hence $F_{g(x)}\times G_x$ is a neighbourhood of $(f,g)$, and

$$\phi(F_{g(x)}\times G_x) \subset \{ h \in S(X,X) : h(x) = f(g(x))\},$$

so $\phi$ is continuous in $(f,g)$.