Is everything right in this set-theory problem? I've got a following homework to solve:
$f:\Bbb{N}^\Bbb{N}\rightarrow \mathcal P(\Bbb{N})$ is such a function that $f(\phi)=\phi(\Bbb{N}). $ Is $f$ bijective? Find $f^{-1}(B)$ where $B$ is a set of one-element subsets of $\Bbb{N}$.
So, according to that description $f(\phi)$ is a subset of $\Bbb{N}$ (an element of $\mathcal P(\Bbb{N})$). But $\phi$ is an element of $\Bbb{N}^\Bbb{N}$ so it should be a function with $\Bbb{N}$ as domain and range. So shouldn't $\phi(\Bbb{N})$ be a natural number? But then how is it possible that $f(\phi)=\phi(\Bbb{N})$? All in all, we have a subset of $\Bbb{N}$ on the left sight of the equation and a natural number on the right. Is my logic correct?
 A: Consider the function
$$
\phi:\mathbb{N}\to\mathbb{N}: \phi(k) = \left\{
\begin{align}
2 & \mbox{ for } k = 1 \\
1 & \mbox{ for } k = 2 \\
k & \mbox{ else}
\end{align}
\right.
$$
and the identity function. Both are different arguments of $f$ but are mapped to the same value $\mathbb{N}$, thus $f$ is not injective and thus not bijective.
For $b \in \mathbb{N}$ we have
$$
f^{-1}(\{b\})= \phi_b \quad \phi_b:\phi_b(k) = b
$$
so the reverse image of a one element set is a constant function of that value.
Assume
$$
B = \{ \{ b_1 \}, \{ b_2 \}, \ldots \} \subset \mathcal{P}(\mathbb{N})
$$ 
then
$$
f^{-1}(B) = \{ \phi \in \mathbb{N}^\mathbb{N} \,|\, f(\phi) \in B \} = \{ \phi_{b_1}, \phi_{b_2}, \ldots \}
$$
A: No! $\phi(\mathbb N)$ is the image of $\phi$, that is a subset of $\mathbb N$.
$f$ can't be bijective since there are functions $\phi,\psi$ with the same image in 
$\mathbb N$. 
If $B=\{n\}$ then $f^{-1}(B)$ is all functions $h:\mathbb N\rightarrow\mathbb N$ for which  $h(m)=n, \forall m$, that is $f^{-1}(\{n\})=\{(0,n),(1,n),(2,n),...\}$ .
