Continuity of a function and its consequences

Question

Let f : X → Y be a given function, and suppose that $f^{−1}(C)$ is an open subset of X whenever C is an open subset of Y .

Prove that f is continuous on X.

Answer 1

Let $f:X\to Y$ so that the preimage of every open subset of $Y$ is open. Let $\varepsilon>0$.

Let $x_0\in X$ and $B_\varepsilon(y_0)=\{y\in Y: d(y_0,y)<\varepsilon\}$ (which is open) where $y_0=f(x_0)\in Y$. Now, $V=f^{-1}(B_\varepsilon(y_0))$ is open, because it is a preimage of an open set.

Because $x_0\in V$, there exsits a $\delta>0$ so that $B_\delta(x_0)\subset V$ (because $V$ is open), which implies $f(B_\delta(x_0))\subset f(V)=B_\varepsilon(y_0)$.

This ($f(B_\delta(x_0))\subset B_\varepsilon(y_0)$) is exactly the $\varepsilon$-$\delta$ definition of continuity at $x_0\in X$, which we chose arbitrarily; so $f$ is continuous on $X$.