How $f(z)=z^n$ is continuous but its inverse doesn't map an open set to an open set?

Question

If a map $f: X \to Y$ is continuous then $f^{-1}(V)$ is open set in $X$ whenever $V$ is an open set in $Y$.

Consider $f: \mathbb{C} \to \mathbb{C}$ by $f(z)=z^n$. By this map the sector $r≤a$, $0≤θ<2π/n$ is mapped onto the disk $ρ ≤ a^n $; and by letting $a \to \infty$, $\mathbb{C}$ is open in $\mathbb{C}$ but $V=\{z \in \mathbb{C} : 0≤θ<2π/n\}$ is not an open set?

Is this contradicts the theorem stated in first line?

Answer 1

If $V \subset X$, then in general $V \neq f^{-1}(f(V))$. We can only say that $V \subset f^{-1}(f(V))$.

In your case, indeed, $V$ is not open, and $f(V) = \mathbb C$ is. But then $f^{-1}(f(V)) = f^{-1}(\mathbb C) = \mathbb C$ (because any complex number has at least one $n$-th root), which is open.