A function $f :X \to Y$ between topological spaces $X,Y$ is defined to be
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continuous if $f^{-1}(V)$ is open in $X$ for all open $V \subset Y$,
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open if $f(U)$ is open in $Y$ for all open $U \subset X$,
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closed if $f(C)$ is closed in $Y$ for all closed $C \subset X$.
Many questions in math.stackexchange deal with alternative characterizations of these properties in terms of the behavior of the operators $\operatorname{int}$ (interior) and $\operatorname{cl}$ (closure) with respect to $f$ and $f^{-1}$. A recent example is Characterization of Continuous, Closed and Open maps. In virtually all textbooks on general topology one can find theorems about such characterizations, but it seems that a complete list is not available. There are eight properties which can be formulated in terms of interior and closure:
| For all $A \subset X$ resp. $B \subset Y$ | |
|---|---|
| (1) | $f(\text{int} A) \subset \text{int} f(A)$ |
| (2) | $f(\text{int} A) \supset \text{int} f(A)$ |
| (3) | $f^{-1}(\text{int} B) \subset \text{int} f^{-1}(B)$ |
| (4) | $f^{-1}(\text{int} B) \supset \text{int} f^{-1}(B)$ |
| (5) | $f(\operatorname{cl} A) \subset \operatorname{cl} f(A)$ |
| (6) | $f(\operatorname{cl} A) \supset \operatorname{cl} f(A)$ . |
| (7) | $f^{-1}(\operatorname{cl} B) \subset \operatorname{cl} f^{-1}(B)$ |
| (8) | $f^{-1}(\operatorname{cl} B) \supset \operatorname{cl} f^{-1}(B)$. |
Question: Try to identify each of these properties with $f$ being continuous, open or closed.
Best Answer
Here it is:
Let us first collect some well-known facts about images and preimages of functions $f : X \to Y$ between sets $X,Y$.
$f^{-1}$ is compatible with all standard set-theoretic operations. That is, $f^{-1}(B \cup B') = f^{-1}(B) \cup f^{-1}(B')$, $f^{-1}(B \cap B') = f^{-1}(B) \cap f^{-1}(B')$, $f^{-1}(Y \setminus B) = X \setminus f^{-1}(B)$. The latter is also written as $f^{-1}(\complement B) = \complement f^{-1}(B)$, where $\complement$ denotes the complement.
$f$ is compatible with union. That is, $f(A \cup A') = f(A) \cup f(A')$.
$A \subset f^{-1}(f(A))$. Equality holds if $f$ is injective.
$f(f^{-1}(B)) \subset B$. Equality holds if and only $B \subset f(X)$.
$A \subset f^{-1}(B)$ if and only if $f(A) \subset B$.
$A \cap f^{-1}(B) = \emptyset$ if and only if $f(A) \cap B = \emptyset$.
Moreover, we shall need the following well-known relation between the operators $\operatorname{int}$ and $\operatorname{cl}$ in a topological space $Z$.
A subset $M \subset Z$ is open if and only if $\complement M$ is closed. The equivalence follows from $f^{-1}(B) = f^{-1} (\complement \complement B) = \complement f^{-1} (\complement B)$.
$f^{-1}(\operatorname{cl}f(A))$ is closed and contains $A$. Hence $\operatorname{cl}{A} \subset f^{-1}(\operatorname{cl}{f(A)})$ which implies $f(\operatorname{cl}{A}) \subset f(f^{-1}(\operatorname{cl}{f(A)})) \subset \operatorname{cl}{f(A)}$.
We have $f(\operatorname{cl}{f^{-1}(B)}) \subset \operatorname{cl}{f(f^{-1}(B)))} \subset \operatorname{cl}{B}$, hence $\operatorname{cl}{f^{-1}(B)} \subset f^{-1}(\operatorname{cl}{B})$.
We have $\operatorname{cl}{f^{-1}(\complement B)} \subset f^{-1}(\operatorname{cl}{\complement B})$. Therefore $$f^{-1}(\operatorname{int} B) = f^{-1}(\complement\operatorname{cl}{\complement B}) = \complement f^{-1}(\operatorname{cl}{\complement B}) \subset \complement \operatorname{cl}{f^{-1}(\complement B)} = \complement \operatorname{cl}{\complement f^{-1}(B)} = \operatorname{int} f^{-1}(B) .$$
For any open $V \subset Y$ we have $f^{-1}(V) = f^{-1}(\operatorname{int} V) \subset \operatorname{int} f^{-1}(V)$, hence $f^{-1}(V) = \operatorname{int} f^{-1}(V)$ which is open.
We have $f(\operatorname{int} A) \subset f(A)$. Since $f(\operatorname{int} A)$ is open, we conclude $f(\operatorname{int} A) \subset \operatorname{int} f(A)$.
For any open $U \subset X$ we have $f(U) = f(\operatorname{int} U) \subset \operatorname{int} f(U)$, hence $f(U) = \operatorname{int} f(U)$ which is open.
Let $x \in f^{-1}(\operatorname{cl}{B})$. Then $f(x) \in \operatorname{cl}{B}$. Let $U$ be an open neigborhood of $x$ in $X$. Then $f(U)$ is an open neigborhood of $f(x)$ in $Y$, hence $f(U) \cap B \ne \emptyset$. This implies $U \cap f^{-1}(B) \ne \emptyset$. We conclude $x \in \operatorname{cl}{f^{-1}(B)}$.
Let $U \subset X$ be open. Since $f^{-1}(\operatorname{cl}{\complement f(U)}) \subset \operatorname{cl}{f^{-1}(\complement f(U))} = \operatorname{cl}{\complement f^{-1}(f(U))} \subset \operatorname{cl}{\complement U} = \complement U$, we get $f^{-1}( \operatorname{cl}{\complement f(U)}) \cap U = \emptyset$ which is equivalent to $ \operatorname{cl}{\complement f(U)} \cap f(U) = \emptyset$. This shows $\operatorname{cl}{\complement f(U)} \subset \complement f(U)$. Therefore $\operatorname{cl}{\complement f(U)} = \complement f(U)$, i.e $\complement f(U)$ is closed and $f(U)$ is open.
We have $f^{-1}(\operatorname{cl}{\complement B}) \subset \operatorname{cl}{f^{-1}(\complement B)}$. Therefore $$\operatorname{int} f^{-1}(B) = \complement \operatorname{cl}{\complement f^{-1}(B)} = \complement \operatorname{cl}{f^{-1}(\complement B)} \supset \complement f^{-1}(\operatorname{cl}{\complement B}) = \complement f^{-1}(\complement \operatorname{int} B) = \complement \complement f^{-1}( \operatorname{int} B) \\= f^{-1}( \operatorname{int} B) .$$
We have $f^{-1}(\operatorname{int}{\complement B}) \supset \operatorname{int}{f^{-1}(\complement B)}$. Therefore $$\operatorname{cl} f^{-1}(B) = \complement \operatorname{int}{\complement f^{-1}(B)} = \complement \operatorname{int}{f^{-1}(\complement B)} \subset \complement f^{-1}(\operatorname{int}{\complement B}) = \complement f^{-1}(\complement \operatorname{cl} B) = \complement \complement f^{-1}( \operatorname{cl} B) \\= f^{-1}( \operatorname{cl} B) .$$
We have $f(A) \subset f(\operatorname{cl}{A})$ which is closed. Hence $\operatorname{cl}{f(A)} \subset f(\operatorname{cl}{A})$.
For any closed $C \subset X$ we have $f(C) = f(\operatorname{cl}{C}) \supset \operatorname{cl}{f(C)}$, hence $f(C) = \operatorname{cl}{f(C)}$ which is closed.
Example 1. $f : \mathbb{R} \to \mathbb{R}^2, f(x) = \begin{cases} (x,0) & x \le 0 \\ (x,1) & x > 0 \end{cases} $ satisfies (2) because $\operatorname{int} f(A) = \emptyset$ for all $A$. $f$ is not continuous, not open and not closed.
Note that $f$ is injective, not surjective and $f(\mathbb{R})$ is not open in $\mathbb{R}^2$.
Example 2. $f: I \times I \to I, f(s,t) = s$ is continuous, open, closed. It does not satisfy (2) because for the diagonal $\Delta \subset I \times I$ we have $f(\operatorname{int}\Delta)) = f(\emptyset) = \emptyset$, but $\operatorname{int} f(\Delta) = \operatorname{int} I = I$.
Note that $f$ is surjective and not injective.
Theorem. Let $f(X)$ be open in $Y$. If (2) is satisfied, then $f$ is continuous.
Proof. Let $V \subset Y$ be open. We have to show that $f^{-1}(V)$ is open in $X$. Let $V' = V \cap f(X)$ which is again open in $Y$. We have $f^{-1}(V) = f^{-1}(V')$. Thus we may assume that $V \subset f(X)$. This is essential for the proof because $f(f^{-1}(M)) = M$ if and only if $M \subset f(X)$.
It is a bit tricky to find a set to which we apply (2).
To prove that $f^{-1}(V)$ is open we show that the difference $D = f^{-1}(V) \setminus \operatorname{int} f^{-1}(V)$ is empty. Let $E = D \cup f^{-1}(V \setminus f(D))$. We have $E \subset f^{-1}(V)$, thus $\operatorname{int} E \subset \operatorname{int} f^{-1}(V)$, in particular $\operatorname{int} E \cap D = \emptyset$. We conclude $$\operatorname{int} E \subset f^{-1}(V \setminus f(D)) .$$ Noting $f(D) \subset V$ we get $$f(E) = f(D) \cup f(f^{-1}(V \setminus f(D))) = f(D) \cup( V \setminus f(D)) = V .$$ But by (2) we get $$V = \operatorname{int} V = \operatorname{int} f(E) \subset f(\operatorname{int} E) \subset f(f^{-1}(V \setminus f(D))) = V \setminus f(D)$$ which is only possible when $f(D) = \emptyset$, i.e. $D = \emptyset$.
Remark. The assumption ''$f(X)$ open in $Y$'' can be relaxed a little. In fact, it suffices to assume that for any open $V \subset Y$, $f(X) \subset V$ or $V' = V \cap f(X)$ is open in $Y$.
Corollary 1. Let $f$ be surjective. If (2) is satisfied, then $f$ is continuous.
Corollary 2. Let $f$ be open. If (2) is satisfied, then $f$ is continuous.
The converse of the above theorem is false (see example 2). But again there is an interesting special case.
Lemma. Let $f$ be injective. If $f$ is continuous, then (2) is satisfied.
Proof. By (3) we have $$f^{-1}(\operatorname{int} f(A)) \subset \operatorname{int} f^{-1}(f(A)) = \operatorname{int} A$$ which implies $$\operatorname{int} f(A) \subset f(\operatorname{int} A) .$$