Homeomorphism between $\mathbb N^\infty$ and a closed subset $\mathbb M$ of $(\mathbb N^\infty)^\infty$

Question

Trying to figure out a proof of a lemma that I'm reading in Stochastic Relations by Ernst-Erich Doberkat.

The Baire space, denoted $\mathbb{N}^\infty$, is the infinite product of the natural numbers. The topology is the infinite product topology of the discrete one, which has as a basis the cylinder sets.

In the Lemma, I'm given a Polish space $X$, and $\omega$ continuous maps $f_n : \mathbb N^\infty \rightarrow X$, and every image $f_n[\mathbb N^\infty]$ is a Borel set.

The goal is to show that there is a continuous map $f : \mathbb N^\infty \rightarrow X$ with $f[\mathbb N^\infty] = \bigcap_{n \in \mathbb N} f_n[\mathbb N^\infty]$ (showing that a collection of "good" sets is closed under countable intersection).

The author proceeds as follows:

Put $\mathbb M := \{ (\tau_1,\tau_2,\ldots) \mid f_1(\tau_1) = f_2(\tau_2) = \ldots \}$, which can be shown to be closed in $(\mathbb N^\infty)^\infty$ by taking a point outside of it, choosing $i<j$ such that $f_i^{-1}(\tau_i) \neq f_j^{-1}(\tau_j)$, and straightforwardly constructing an open set that does not intersect $\mathbb M$.

He then defines a map $f : \mathbb M \rightarrow X, (\tau_1,\tau_2,\ldots) \mapsto f_1(\tau_1)$, which is continuous, and has the desired property that $f[\mathbb M] = \bigcap_{n\in \mathbb N} f_n[\mathbb N^\infty]$.

He then very casually states that $\mathbb M$ is homeomorphic to $\mathbb N^\infty$ and concludes the proof.

I have two questions:

1. I get that if we have a continuous surjective function $g : \mathbb N^\infty \rightarrow \mathbb M$, then we have $f \circ g$ continuous and $(f \circ g)[\mathbb N^\infty] = f[\mathbb M]$, which is what we need, so do you agree that $g$ need not be a homeomorphism?
2. Why is $\mathbb M$ homeomorphic to $\mathbb N^\infty$?

EDIT: as pointed out by Mateo in the comments, $\mathbb{M}$ cannot be homeomorphic to $\mathbb N^\infty$, as we can easily think of continuous functions $f_n : \mathbb N^\infty \rightarrow X$ whose images have empty intersection, making sure that $\mathbb M = \emptyset$. (Take a bunch of constant functions, for example).

The author must have made some kind of mistake.

The question remains whether we can find continuous $g:\mathbb N^\infty \rightarrow (\mathbb N^\infty)^\infty$ such that $g[\mathbb N^\infty] = \mathbb M$ (or, equivalently, continuous surjective $g : \mathbb N^\infty \rightarrow \mathbb M$)