## Baire space homeomorphism question - again

I hope someone will read this long question to the end, I'm getting really desperate.

Here's what I have to prove:
Let $\psi: \mathbb{N}^d \times B^k \rightarrow B$ be defined as follows:
$\alpha=(n_1, \ldots, n_d, f_1, \ldots, f_k)$
$\psi(\alpha)=(n_1, \ldots, n_d, f_1(0), f_2(0), \ldots, f_k(0), f_1(1), \ldots, f_k(1), \ldots)$,
where B is the Polish set $B=\mathbb{N}^{\mathbb{N}}$ with product topology and every $\mathbb{N}$ has discrete topology.
Show that $\psi$ is a homeomorphism.

Injectivity and surjectivity is easy, continuity isn't.

Here are the facts:
$f:B \rightarrow B$ is continuous if $\forall x \in B$ and $\forall \sigma \subset f(x)$ $\exists \tau \subset x$ such that $\tau \subset y \Rightarrow \sigma \subset f(y)$.

$f:X\rightarrow \Pi Y_i$ is continuous if and only if $\pi_i \circ f$ is continuous, for all i.

Let $\phi=\psi^{-1}$ be the inverse.
$\phi: B \rightarrow \mathbb{N}^d \times B^k$ is continuous if $p \circ \phi$ and $q \circ \phi$ are continuous, where
$p: \mathbb{N}^d \times B^k \rightarrow \mathbb{N}^d$ takes $(x, y) \in \mathbb{N}^d \times B^k$ to $x\in \mathbb{N}^d$.
$q: \mathbb{N}^d \times B^k \rightarrow B^k$, akes $(x, y) \in \mathbb{N}^d \times B^k$ to $y\in B^k$.

Since $\mathbb{N}^d$ and $B^k$ are also products, we can say that $\phi$ is continuous if $p_i \circ \phi$, $i=1, \ldots, d$ and $q_j \circ \phi$, $j=1,\ldots, k$ are continuous.
$p_i$ sends $(x, y)$ to the i-th coordinate of the $\mathbb{N}^d$part, and $q_j$ sends $(x, y)$ to the j-th coordinate of the $B^k$ part.

Now it's easy to prove that the inverse is continuous.

The question is: what do I do about $\psi$?