Is there a continuous surjective map from $\mathbb{R}$ onto the subspace $\mathbb{R^{\infty}}$ of $\mathbb{R}^\omega$ with product topology, where $\mathbb{R^{\infty}}$denotes the set of points in $\mathbb{R}^\omega$such that all but finitely many coordinates of the points are zero.
What if $\mathbb{R}^\omega$ is equipped with box topology or uniform topology?
Define the subsets $S_n$ of $\mathbb R^\infty$ by $S_n = \big\{ \{a_1,\dots,a_n,0,0,\dots\} \colon \text{each }|a_j|\le n \big\}$. Suppose $f\colon \mathbb R \to \mathbb R^\infty$ is a map with the following property: for every $n$, there exists an interval $I_n\subset\mathbb R$ such that $f$ restricted to $I_n$ is a surjection onto $S_n$. Then $f$ is a surjection onto $\mathbb R^\infty$.
It's easy to see that this implication is true, since every point in $\mathbb R^\infty$ is in $S_n$ for all sufficiently large $n$ (all we need is that every point is in one $S_n$). It's also easy to construct such a function $f$, by making it implement a continuous space-filling curve onto $S_n$ on the interval $[2n,2n+1]$ and then any old thing on the interval $[2n+1,2n+2]$ to keep it continuous.
One would have to work out how sensitive this construction is to the chosen topology - it reduces to whether there are continuous space-filling curves in those topologies.
