Continuity on Box Topology I was reading about box topology and product topology from Munkres's Topology. Their is an example given below:
$$\mathbb{R}^{\omega}=\prod_{n\in \mathbb{Z}_+}X_n$$
where $X_n=\mathbb{R}$ for each $n$.
$$f: \mathbb{R}\rightarrow \mathbb{R}^{\omega}$$
$$f(t)=(t,t,\dots), t\in \mathbb{R}$$
the $n$-th coordinate function of $f$ is the function $f_n(t)=t$. It is clear by the counter example given that if $\mathbb{R}^{\omega}$ is box topology then $f$ is not continuous.
My question:
$$\mathbb{R}^{\omega}=\prod_{n\in \mathbb{Z}_+}X_n$$
where $X_n=\mathbb{R}$ for each $n$.
$f: \mathbb{R}\rightarrow \mathbb{R}^{\omega}$. The $n$-th coordinate function of $f$ is the function $f_n:\mathbb{R}\rightarrow \mathbb{R}$ is a non-constant onto continuous function, $\mathbb{R}^{\omega}$ is box topology.
Then can $f$ ever be continuous? What if we omit onto?
 A: Yes, such an $f$ can be continuous.  To prove this, note that any function to $\mathbb{R}^\omega$ which is continuous on each coordinate and is constant on all but finitely many coordinates is continuous with respect to the box topology.  So consider $f:\mathbb{R}\to\mathbb{R}^\omega$ such that for each $n\in\mathbb{Z}$, $f$ is constant on all but one of the coordinates when restricted to $[n,n+1]$ (and continuous on all coordinates).  Since continuity is local, such an $f$ is continuous with respect to the box topology.  By changing which coordinate $f$ is non-constant on for different integers $n$, we can arrange that each coordinate function of $f$ is surjective. There are only countably many things we need to arrange this way (we need to make sure for each coordinate $i$ and each $N\in\mathbb{N}$, the image of $f_i$ contains $[-N,N]$), and so we can use one interval $[n,n+1]$ for each one of them.
Note that this is essentially the only way to get such an example.  To be more precise, if $f:\mathbb{R}\to\mathbb{R}^\omega$ is continuous with respect to the box topology, then each $t\in\mathbb{R}$ has a neighborhood on which $f$ is constant on all but finitely many coordinates.  Indeed, suppose that for all $\epsilon>0$, there are infinitely many coordinates on which $f$ is nonconstant on $(t-\epsilon,t+\epsilon)$.  For each $n$, choose a coordinate $k_n$ and $\delta_n>0$ such that there exists $s\in (t-1/n,t+1/n)$ such that $|f_{k_n}(s)-f_{k_n}(t)|<\delta_n$.  Furthermore, make these choices so that the $k_n$ are all distinct (we can do this since for each $n$, there are infinitely many coordinates to choose, so we can pick one which is not $k_m$ for any $m<n$).  The set $U$ of points $(x_i)$ such that $|x_{k_n}-f_{k_n}(t)|<\delta_n$ for all $n$ is then an open neighborhood of $f(t)$ in the box topology.  By construction, $f^{-1}(U)$ does not contain $(t-1/n,t+1/n)$ for any $n$, so it is not open.  Thus $f$ is not continuous.
