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The usual definition of a continuous map between two topological spaces is that a map is continuous if the preimage of every open set is open.
I believe, but am not sure, that to prove a map is continuous it suffices to show that the preimage of every closed set is closed. Or perhaps this only works if the map is surjective ...
Is this true?
It's equivalent. One can argue as follows, where $\bullet$ and $\circ$ may be replaced by "open" and "closed" in either order:
Given any topological spaces $X$ and $Y$, and function $f:X\to Y$, suppose that for every $\bullet$ set $C\subseteq Y$, $f^{-1}(C)$ is a $\bullet$ subset of $X$.
Note that $C$ is $\bullet$ if and only if $\overline{C}$, the complement of $C$, is $\circ$. Since $f^{-1}(C)$ is $\bullet$, we must have that $$f^{-1}(\overline{C})=\{x\in X\mid f(x)\notin C\}=\{x\in X\mid x\notin f^{-1}(C)\}=\overline{f^{-1}(C)}$$ is $\circ$. Thus, for every $\circ$ subset $B\subseteq Y$, the preimage $f^{-1}(B)$ is also $\circ$.
The reason that we are able to use closed sets in the definition of continuity is that complimentation "behaves well" under taking preimages. What I mean by this is that $f^{-1}(A^{c})=(f^{-1}(A))^{c}$ for all functions $f:X\to Y$ and sets $A\subset Y$. Since the compliment of any open set is closed, if we take $A$ to be an open set then $A^{c}$ is closed. If we consider our new definition of continuity, we must have that $f^{-1}(A^{c})$ is closed if $f$ is continuous. But we have that $f^{-1}(A^{c})$ is closed from our assumption, and so its compliment must be open. But the compliment of this set is just $f^{-1}(A)$, and so we have shown that the inverse image of every open set is open. This is just the standard definition for continuity. Hence, we have shown that (closed) continuity implies the standard continuity. The other implication has exactly the same proof.
In fact, the set of all open sets in $\mathbb{R}^{n}$ (or any metric space) is a topology. However, if we take a base for the topology (open balls in the metric space for example), then if we have the inverse image of every base set is open, then the inverse image of every set in the topology must also be open. The proof is not too hard because it follows from the definition of a basis.
I hope this post helped!
Yes this is true. In fact the following are equivalent for a function $f:X\rightarrow Y$:
1) $f$ is continuous.
2) If $O$ is open in $Y$, then $f^{-1}(O)$ is open in $X$.
3) If $K$ is closed in $Y$, then $f^{-1}(K)$ is closed in $X$.
4) $f( CL_X (E)) \subset CL_Y(f(E))$; where $CL_X(E)$ is the closure of $E$ in $X$.
See, e.g., page 44 in Willard's General Topology.
