Define the function $f: \mathbb{N} \to Y$ where $\mathbb{N}$ has the discrete topology, and $Y$ is the set $\{0, \frac{1}{n} | n \in \mathbb{N} \}$ with the relative topology from $\mathbb{R}$:
$$ f(n)= \begin{cases} 0 & n=1\\ \frac{1}{n - 1} & n>1\\ \end{cases} $$
Is $f$ continuous? Is $f^{-1}$ continuous?
My attempt:
$f$ is continuous since $\mathbb{N}$ has the discrete topology.
$f^{-1}$ is not continuous. If we consider the open set $U = \{1, 3, 5, 7, ... \}$, then $f(U) = \{0, \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, ... \}$. But $f(U)$ is not an open set in $Y$ since any open set containing $0$ must also contain elements of the form $\frac{1}{n}$ where $n$ is odd.
Is my reasoning right?
Your right about $f$: any function on a discrete space is continuous. But more simply: $f^{-1}$ is not continuous as $f$ is not open: $f[\{1\}] = \{0\}$ and the latter is not open as any neighbourhood of $0$ also contains infinitely many points of the form $\frac{1}{n}$.