Show that $F$ is continuous by showing that the component functions are continuous in the product topology. 
Let $F : \mathbb{R}^\mathbb{R} \to \mathbb{R}^\mathbb{R}$ be a map such that $F(f)(x) = g(x)f(h(x))$ for fixed $g$ and $h$ maps from $\Bbb R \to \Bbb R$. Show that $F$ is continuous by showing that the component functions are continuous in the product topology.

My idea was to pick a basis set $U = \bigcap_{x \in K} \pi_x^{-1}(U_x)$ where $K \subset \Bbb R$ is finite and show that the preimage is open. What I get is that $$F^{-1}(U) = F^{-1}\left( \bigcap_{x \in K} \pi_x^{-1}(U_x) \right) = \bigcap_{x \in K} F^{-1}(\pi_x^{-1}(U_x)) = \bigcap_{x \in K} ( \pi_x \circ F)^{-1}(U_x).$$
From here I wanted to figure out whether $\pi_x \circ F$ is open i.e. the $x$'th projection, but I don't understand what are these projections? In the product topology I think that $\pi_x(F) = F(x)$, but it's not helpful in my case. What can be done with this?
 A: Take $x_0\in\Bbb R$ and an open subset $A$ of $\Bbb R$. Consider the set$$O(x_0,A)=\left\{\varphi\in\Bbb R^{\Bbb R}\,\middle|\,\varphi(x_0)\in A\right\},$$which is an open subset of $\Bbb R^{\Bbb R}$. I will compute $F^{-1}\bigl(O(x_0,A)\bigr)$ and I will prove that it is an open subset of $\Bbb R^{\Bbb R}$ too. There are three cases to be taken into account:

*

*$g(x_0)=0$ and $0\in A$: then $F(f)\in O(x_0,A)\iff g(x_0)f\bigl(h(x_0)\bigr)(=0)\in A$, which holds for any $f$. So, $F^{-1}\bigl(O(x_0,A)\bigr)=\Bbb R^{\Bbb R}$, which is an open set.

*$g(x_0)=0$ and $0\notin A$: then $F(f)\in O(x_0,A)\iff g(x_0)f\bigl(h(x_0)\bigr)(=0)\in A$, which never holds. So, $F^{-1}\bigl(O(x_0,A)\bigr)=\emptyset$, which is an open set.

*$g(x_0)\ne0$: then,\begin{align}F(f)\in O(x_0,A)&\iff g(x_0)f\bigl(h(x_0)\bigr)\in A\\&\iff f\bigl(h(x_0)\bigr)\in\frac1{g(x_0)}A\\&\iff f\in O\left(h(x_0),\frac1{g(x_0)}A\right),\end{align}which is an open set.

Since the sets of the form $O(x_0,A)$ span the topology of $\Bbb R^{\Bbb R}$ and since the inverse image of each such set (with respect to $F$) is an open set, $F$ is continuous.

Another possibility is to use the fact that $F$ is continuous if and only if, for any $x\in\Bbb R$, $\pi_x\circ F$ is continuous. But, if $x\in\Bbb R$ and $f\in\Bbb R^{\Bbb R}$,$$\pi_x\bigl(F(f)\bigr)=g(x)f\bigl(F(h(x)\bigr)=g(x)\cdot\pi_{h(x)}(f).$$Since $\pi_{h(x)}$ is continuous and $g(x)$ is constant, $\pi_x\circ F(=g(x)\cdot\pi_{h(x)})$ is continuous.
