Mapping is Continuous iff Inverse Images of Open Sets are Open/Corollary

Theorem
Let $X$ and $Y$ be normed vector spaces.

Let $f : X \to Y$ be a mapping.

Then:
 * $f$ is continuous on $X$


 * for every $F$ closed in $Y$, $\map {f^{-1}} F$ is closed in $X$.
 * for every $F$ closed in $Y$, $\map {f^{-1}} F$ is closed in $X$.

Sufficient Condition
Suppose that for every closed $F$ in $Y$, $\map {f^{-1}} F$ is closed in $X$.

Let $V$ be open in $Y$.

By definition, $Y \setminus V$ is closed in $Y$.

By assumption, $\map {f^{-1}} {Y \setminus V}$ is closed in $X$.

We have that:

Thus, $X \setminus \map {f^{-1} } V$ is closed in $X$.

By definition, $X \setminus \map {f^{-1} } V$ is closed iff $\map {f^{-1} } V$ is open.

Hence, for every $V$ open in $Y$, $ \map {f^{-1}} V$ is open in $X$.

By Mapping is Continuous iff Inverse Images of Open Sets are Open, $f$ is continuous on $X$.

Necessary Condition
Suppose that $f$ is continuous on $X$.

Let $F$ be closed in $Y$.

By definition, $Y \setminus F$ is open in $Y$.

By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {f^{-1} } {Y \setminus F}$ is open on $X$.

Hence:

and $X \setminus \map {f^{-1} } F$ is open in $X$.

By definition, $\map {f^{-1}} F$ is closed in $X$