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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 $B \subset Y$

Seth
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  • If $2$ is from Engelking (as I think it is: 1.4.C): note that open mapping means continuous and open simultaneously here, as defined on the bottom of page 31, where a mapping is only called open if it is continuous first (it's a class of continuous maps as E puts it). – Henno Brandsma Sep 12 '19 at 21:14
  • And 2 should have $=$ instead of $\subset$ for the interior and note that in both cases one inclusion is equivalent to continuity (1.4.1 sub v' and vi) and the other one is exactly equivalent to open-ness. The equivalence follows from the usual formula that converts interiors to closures e.g. – Henno Brandsma Sep 12 '19 at 21:16

1 Answers1

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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.

Henno Brandsma
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  • Thank you for showing me how to do question 1. I was really struggling to solve it as well as how to show $f^{-1](\overline{B}) \subset \overline{f^{-1}(B)}$. I thought if i can solve the latter, it will help me with question 1. But now I will give number 2 a try again. – Seth Sep 12 '19 at 08:18
  • if you have a moment, I was wondering if you can please clarify something for me. :Different topology texts give different theorems about open and closed maps. I just want to be clear in what contexts and under what conditions do open and closed maps exists. Here is the link https://math.stackexchange.com/questions/3353579/questions-concerning-properties-of-open-and-closed-maps-in-general-topological-s – Seth Sep 12 '19 at 08:20
  • @SethMai open mapping and open function can be different things. Be careful and read all the preceding definitions in a book or text as well. – Henno Brandsma Sep 12 '19 at 08:40
  • @SethMai so in 2 you can and must use that $f$ is continuous too, or it will be false. 1 holds for just open functions. – Henno Brandsma Sep 12 '19 at 08:42
  • both statements in 2, I can assume that f is continuous. What if I just assume f to be only one to one? – Seth Sep 12 '19 at 08:46
  • @SethMai no you may not assume it’s 1-1 but $f$ is given to be a mapping so is continuous. – Henno Brandsma Sep 12 '19 at 08:47
  • thank you for clearing that up for me. – Seth Sep 12 '19 at 08:50
  • Let us continue this discussion in chat. – Seth Sep 12 '19 at 09:30
  • Here is 1, I also messaged you.

    we need to show that $f(A)$ is open set in $Y$. Meaning, for any $W \subset X$, and $y \in W$, $W(y) \subset f(A)$. Let $f(A)=B$ and $A=f^{-1}(B)$, from hypothesis, x is in $\partial f^{-1}(B)$, hence $x \in \overline{f^{-1}(B)} \cap (X-f^{-1}(B)$). Every neighborhood $O$, where $O \subset X$ of x has non intersection with $f^{-1}(B)$. $O(x) \subset f^{-1}(B)$ implies $A$ is open and for some $y=f(x)\in f(O)$ and we can let $W(y)=f(O)$, $f(x) \in f(O) \subset B$, and $f(A)=B$, we have $f(O) \subset f(A)$ Hence, f is an open function

     – Seth Sep 12 '19 at 09:50
  • @SethMai Your idea for the reverse doesn't make sense. Have a look at my effort instead. $W(y)$? I don't see how you propose to show openness of $f[O]$ in your "proof". – Henno Brandsma Sep 12 '19 at 19:16
  • Henno, My apologies for not responding earlier to your replies here and my other SE posts. I have not been feeling well and have been away from the computer... – Seth Sep 15 '19 at 00:26
  • ...After i posted my attempted solution, i realized i made some mistakes. I was trying to do something to what you did by trying with a direct proof. So if I have an element in the the boundary of the inverse image, and I have an open set from the domain. I can hopefully work towards showing that $f(A)$ is open by using what it means for an element to be in the boundary of a set and proceed from there. It is similar to what graydad did here:.... – Seth Sep 15 '19 at 00:37
  • https://math.stackexchange.com/questions/1832039/proving-these-equivalent-conditions-for-an-open-map-using-boundary-of-a-set I see that you also post a solution yourself there as well.... The notation $W(y)$, it was to denote an element in an open set, or the neighborhood of a y. I have seen such notation used in other topology texts. However I am not sure what the standard notation for the neighborhood of an element. My apologizes if it was confusing... – Seth Sep 15 '19 at 00:41
  • ...the reason I did not use the prove by contradiction approach because I was not comfortable with what it means for a set to be open. If I can ask here of you to clarify here for me... The standard theorem on conditions for a set to be open is the following: A set $A$ is open iff $A$ is $\mathring A=A$. So if set A is not open, do you have still have the if and only if condition as follows, (1) $A$ is not open if and only if $A \not\subset\mathring A$ Also, (2) $A$ is not open iff $\partial A \not\subset A^{c}$. Actually, does (2) follow from (1) because... – Seth Sep 15 '19 at 01:02
  • ...actually, i don't know how (2) follows from (1). – Seth Sep 15 '19 at 01:38
  • @SethMai (2) does not follow from (1), not right away at least. (logically speaking any true statement follows from another true statement, but here we mean "easily reducible to"). Prove (2) (which is 2 problems, not 1) as its own problem. If you have trouble proving them, start a new question on where you're stuck. – Henno Brandsma Sep 15 '19 at 05:32