Questions about inverse image and boundary of open maps and closed maps in general topological spaces I am having a trouble solving the following two questions. I think question 2 helps with solving question 1.  But since I am not sure what else I can assume from a function that is an open map, I don't know how to proceed.  I tried using various results from closure boundary and interiors of sets as well as properties of direct and inverse images of functions.  I am not having any success.  Thank you in advance.
1)Show that $f:(X,\tau)\rightarrow (Y,\sigma)$ is open if and only if $f^{-1}(\partial B)\subset \partial f^{-1}(B)$, for each $B \subset Y$
2) Show that a mapping $f$ of $X$ to $Y$ is open if and only if $\overline{f^{-1}(B)}=f^{-1}(\overline B)$ or - equivalently - $\mathring{f^{-1}(B)}\subset f^{-1}(\mathring{B})$ for every $B \subset Y$ 
 A: It's a matter of definitions and proving two implications. We do not necessarily have to use identities for boundaries etc. if we don't see a need for them. If in doubt, go to first principles.
E.g. Suppose $f$ is open and let $B \subseteq Y$. We want to show that $f^{-1}[\partial B] \subseteq \partial f^{-1}[B]$. So take a point $x \in f^{-1}[\partial B]$. To see it is in $\partial f^{-1}[B]$, let $O$ be any open neighbourhood of $x$ and we show it intersects both $f^{-1}[B]$ and its complement.
What do we know? $f(x) \in \partial B$ by definition of inverse image. And $f[O]$ is an open neighbourhood of $f(x)$ as we are proving the left-to-right implication of (1) so we know $f$ is open. 
So by the definition of $\partial B$ we know $f[O] \cap B \neq \emptyset$ and $f[O] \cap Y\setminus B \neq \emptyset$. So we know there is some $x_1 \in O$ with $f(x_1) \in B$ and some $x_2 \in O$ with $f(x_2) \notin B$. But then $x_1 \in f^{-1}[B]$ and $x_2 \notin f^{-1}[B]$ (definitions!!). And so $O$ intersects (in these $x_1,x_2$ resp.) both $f^{-1}[B]$ and its complement, ergo $x \in \partial f^{-1}[B]$ as we needed to show.
And no point have I been creative: I have just written down what I had to show by definition and followed the definitions and the only useful fact about $f$ I had: it being open. 

For the reverse, assume now that $$\forall B \subseteq Y: f^{-1}[\partial B] \subseteq \partial f^{-1}[B]\tag{1}$$
and we'll show $f$ is open. So let $O$ be open in $X$ and consider $f[O]$. $f[O]$ is open iff $\partial f[O]$ is disjoint from $f[O]$ (to make a connection with boundaries, where we can use our assumption).
So apply $(1)$ to $B=f[O]$ and we get
$$f^{-1}[\partial f[O]] \subseteq \partial f^{-1}[f[O]]\tag{2}$$
Now suppose we had some point $y \in \partial f[O] \cap f[O]$, so $y =f(x)$ with $x \in O$ and $f(x) \in \partial f[O]$ but then $x$ is in the left hand side of $(2)$ and so $x \in \partial f^{-1}[f[O]]$. But this is a contradiction as $x \in O \subseteq f^{-1}[f[O]]$, and so $x$ is in the interior of $f^{-1}[f[O]]$ not in its boundary. This contradiction shows that $f[O] \cap \partial f[O] = \emptyset$ and so $f[O]$ is open. QED.
This implication is slightly trickier, as we need to know that $O$ is open iff its disjoint from its boundary (as then $\partial O = \partial O^\complement \subseteq O^\complement$ so $O^\complement$ is closed and $O$ is open etc.), and this suggests taking $B=f[O]$ in the assumption and the rest is just reasoning through the consequences. There might be a clean algebraic proof too, but this is what I can come up with.
