$f$ continuous $ \Rightarrow f^{-1} (Int(B)) \subset Int (f^{-1}(B)) \Rightarrow bd (f^{-1}(B)) \subset f^{-1}(bd (B)) \Rightarrow f $ is continuos. Let $X$ and $Y$ be topological spaces and $f:X \rightarrow Y$. Prove that the following conditions are equivalent:
a) $\quad f$ is continuous.
b) $\quad f^{-1} (Int(B)) \subset Int (f^{-1}(B))$ for every $B \subset Y$.
c) $\quad \partial (f^{-1}(B)) \subset f^{-1}(\partial (B))$ for every $B \subset Y$.
I want to show it in the order $ a) \Rightarrow b) \Rightarrow c) \Rightarrow a)$.
I already showed $a) \Rightarrow b)$. For $b) \Rightarrow c)$, Let $x \in \partial (f^{-1}(B))$ and I must prove that $x$ is also in $f^{-1}(\partial (B))$. But I get stuck trying to relate this with the hypothesis.
For $c) \Rightarrow a)$, Let $x \in X$ and $V$ a neighborhood of $f(x)$. It must be found a neighborhood $U$ of $x$ such that $f(U) \subset V$. But also I get stuck.
 A: $(b) \to (a)$ is very easy:$\DeclareMathOperator{\Int}{Int}$
If $V \subseteq Y$ is open, then $f^{-1}(V) = f^{-1}(\Int V) \subseteq \Int f^{-1}(V)$. But, we also have $\Int f^{-1}(V) \subseteq f^{-1}(V)$, since this holds for any set. Hence $\Int f^{-1}(V) = f^{-1}(V)$ so $f^{-1}(V)$ is open.
Therefore, $f$ is continuous.
$(a) \to (c)$
Let $B \subseteq Y$ and $x \in \partial f^{-1}(B)$. We wish to show that $x \in f^{-1}(\partial B)$, that is $f(x) \in \partial B$. Let's show that any open neighbourhood of $f(x)$ intersects both $B$ and $B^c$.
Take an open neighbourhood $V$ of $f(x)$. Then, using $(a)$, we see that $f^{-1}(V)$ is an open neighbourhood of $x$ so it intersects both $f^{-1}(B)$ and $f^{-1}(B)^c$:
$$f^{-1}(V \cap B) = f^{-1}(V) \cap f^{-1}(B) \ne \emptyset \implies V \cap B \ne \emptyset$$
$$f^{-1}(V \cap B^c) = f^{-1}(V) \cap f^{-1}(B^c) = f^{-1}(V) \cap f^{-1}(B)^c \ne \emptyset \implies V \cap B^c \ne \emptyset$$
Since $V$ was arbitrary, $x \in f^{-1}(\partial B)$. We conclude $\partial f^{-1}(B) \subseteq f^{-1}(\partial B)$, which completes the proof.
$(c) \to (b)$
Take $V \subseteq Y$ open. We wish to show that $f^{-1}(V)$ is open in $X$.
From $(b)$ we have:
$$\partial f^{-1}(V^c) \subseteq f^{-1}(\partial V^c) \subseteq f^{-1}(V^c)$$
Since a set is closed if and only if it contains its boundary, we get that $f^{-1}(V^c)$ is closed.
This means that $V = f^{-1}(V^c)^c$ is open. Since $V$ was arbitrary, we conclude that $f$ is continuous.
A: c) to a): Let $O$ be open in $Y$ and assume that $f^{-1}[O]$ is not open in $X$; this must mean that some  point $x \in f^{-1}[O]$ is not an interior point of it. This then says that $x \in \partial f^{-1}[O]$ (every neighbourhood of $x$ intersects $O$ and its complement). But then $x \in f^{-1}[\partial O]$ by assumption c) applied to $B=O$. So $f(x) \in \partial O$. But $x \in f^{-1}[O]$ means $f(x) \in O$ and for open sets $O \cap \partial O = \emptyset$ so contradiction. Hence, $f$ must be continuous.
b) to c) is more awkward I think. If b) holds and $x \in \partial f^{-1}[B]$, we need to show $x \in f^{-1}[\partial B]$ or $f(x) \in \partial B$. Suppose not, we have an open set $O$ containing $f(x)$ that either misses $B$ or $Y \setminus B$. For this $O$ we know by b), using that $O$ is open iff $\operatorname{int}(O) = O$, that $f^{-1}[O] \subseteq \operatorname{int} f^{-1}[O]$, or $f^{-1}[O]$ is open (easy route from b to a, actually). But $O \cap B = \emptyset$ implies $f^{-1}[O] \cap f^{-1}[B] = \emptyset$ and as $x$ is in $f^{-1}[O]$ (as $f(x) \in O$), this would show that $x \notin \partial f^{-1}[B]$, contradiction. We get the same contradiction if $O \cap Y \setminus B = \emptyset$, so our assumption that $f(x) \notin \partial B$ was wrong and we are done.
I think it's slightly easier to do $a \leftrightarrow b$ and $a \leftrightarrow c$ if you study the proofs. Just a bit more work, but more natural, IMHO.
