Questions concerning properties of open and closed maps in general topological spaces

Question

I am confused about several things about open and closed mapping in relations to the listed and question below. I came across them in two different general topology texts. Thank you in advance.

1) Verify that a mapping $f$ of $X$ to $Y$ is closed if and only if $\overline{f(A)}=f(\overline{A})$ for every $A\subset X$ and that $f$ is an open mapping if and only if $f$ is continuous and $f(\mathring{A})\subset \mathring{f(A)}$ [$f(Int A) \subset Int f(A)$]

Here, am I to interpret to mean:

A mapping $f$ of $X$ to $Y$ is closed if and only if $\overline{f(A)}=f(\overline{A})$ for every $A\subset X$
if and only if
$f$ is an open mapping if and only if $f$ is continuous and $f(\mathring{A})\subset \mathring{f(A)}$ [$f(Int A) \subset Int f(A)$]

or are the two separate statements?

Theorem 1 (according to Engelking's General Topology Revised and completed edition): For a one-to-one mapping $f$ of a topological space $X$ onto a topological space $Y$ the following conditions are equivalent:
(i) the mapping $f$ is a homeomorphism.
(ii) The mapping $f$ is closed.
(iii) The mapping $f$ is open.
(iv) The set $f(A)$ is closed in $Y$ if and only if $A$ is closed in $X$
(iv') The set $f^{-1}(B)$ is closed in $X$ if and only if $B$ is closed in $Y$
(v) The set $f(A)$ is open in $Y$ if and only if $A$ is open in $X$.
(v') The set $f^{-1}(B)$ is open in X if and only if $B$ is open in $Y$.

For theorem 1 above, for a function to be open and hence continuous between topological spaces, all it need from the function is that it being one to one.

Theorem 2 (for one direction stated in Dugundji's Topology Text, I changed the notation from $p$ to $f$ for function, otherwise everything else is as worded in the text)

(2b-1) Let $f:X\rightarrow Y$ be a closed map. Given any subset $S \subset Y$ and any open $U$ containing $f^{-1}(S)$, there exists an open $V \supset S$ such that $f^{-1}(V) \subset U$

(2b-2) Let $f:X\rightarrow Y$ be a closed map. Given any subset $S \subset Y$ and any open $A$ containing $f^{-1}(S)$, there exists an open $B \supset S$ such that $f^{-1}(B) \subset A$

Theorem 3 (according to Engelking's General Topology Revised and completed edition) A continuous mapping $f:X\rightarrow Y$ is closed (open) if and only if for every $B\subset Y$ and every open (closed) set $A\subset Y$ which contains $f^{-1}(B),$ there exists an open (a closed set $C\subset Y$ containing $B$ such that $f^{-1}(C)\subset A$

For theorems 2 and 3 above, in one direction for theorem 2 (2b-1 and 2b-2), I don't need to assume anything about the function other than it is open. But if I want to show the converse, which is theorem 3 above, I do need the function to be continuous? What about given theorem 1, can't I just change the function to be just one to one, then i would get the converse?

An open map needn't be continuous: https://math.stackexchange.com/a/75601/403337 — , Sep 12 '19 at 03:04
@ChrisCuster that I know. But here the theorem from the particular text i referenced, it seems that all one need is a much more simpler criteria, then one get everything they want. It seems like a bargain. — Seth, Sep 12 '19 at 03:06
A “mapping” often is a term for “continuous function” and that’s the case in your first theorems. So continuity is assumed throughout. — Henno Brandsma, Sep 12 '19 at 08:38
@HennoBrandsma do you mean when i see a theorem or question about open/closed functions and I see the word "mapping" along with it. I can assume that the function is continuous. But if I don't "mapping" present, and just prove such and such function is either open/closed function or a function's properties, then i can't assume anything about the function. Am I correct? — Seth, Sep 12 '19 at 08:48
@SujitBhattacharyya false. An onto map can be continuous and open and not closed. — Henno Brandsma, Sep 12 '19 at 08:50
@SethMai I mean that you have to look at the whole text and not just cherrypick a statement from it without looking at all definitions involved. — Henno Brandsma, Sep 12 '19 at 08:52
@HennoBrandsma, I look through Engelking's text. He only uses mapping to talk about the definition of a function. That is why these theorems seem to be confusing. It seems like some of them only have continuity as requirement,another one assumes function only being injective. In many instances, the common statement is that if a function is open, then nothing else can be inferred. Here, these theorems are saying if we assume much less, then we can infer even more. So which is which. — Seth, Sep 12 '19 at 08:57
@SethMai the full Engelking statement is true for a continuous $f$ that is also 1-1. Open and closed are equivalent for 1-1 functions in general, but we lose the equivalence with homeomorphism when the function is not continuous. — Henno Brandsma, Sep 12 '19 at 09:01
@HennoBrandsma does that mean if f is either an open nor closed function, then f is also one to one only. But if the function is not continuous, then an open/closed function and being one to one is not equivalent to being homeomorphic. Is that what you meant? — Seth, Sep 12 '19 at 09:06
@SethMai no, I mean that if you know a function is 1-1 and it’s open then it’s closed too, and vice versa. But a function can be open and closed and not 1-1 too. Etc. — Henno Brandsma, Sep 12 '19 at 09:15
@HennoBrandsma but as far the theorem 1 I reference from Engelking in my post. The function f is implicity assume to be continuous. The way is phrased with all the equivalence, the wording looks like given a one to one function, we can show that the function is homeorphic, which implies it is open, then, closed, etc, etc all the way back to showing that it is homeorphic. My apologizes if i am belabouring this. — Seth, Sep 12 '19 at 09:22
@HennoBrandsma or am I to interpret the theorem to be saying, if I am given a one to one function, and the function is also homeomorphic, which is (i), then it is equivalent to (ii), (iii), etc, etc — Seth, Sep 12 '19 at 09:24
@SethMai you’re given a 1-1 continuous function. If you know one of the subsequent facts you know all the others too. — Henno Brandsma, Sep 12 '19 at 09:39
@Mirko Engelking on pg. 1 states: The relation $f \subset X \times Y$ is called a function from $X$ to $Y$, or a mapping of $X$ to $Y$, if every $x \in X$ there exists a $y \in Y$ such that... — Seth, Sep 12 '19 at 16:53
@SethMai page 31 bottom makes clear that an open mapping is in particular a continuous one. — Henno Brandsma, Sep 12 '19 at 21:41

score 1 · Answer 1 · answered Sep 12 '19 at 22:09

The first fact from Engelking means:

Let $f$ be a function from $X$ to $Y$. Then $f$ is continuous and closed iff $f[\overline{A}]= \overline{f[A]}$ for all subsets $A$ of $X$.

The second is separate but similar:

Let $f$ be a function from $X$ to $Y$. Then $f$ is continuous and open (i.e. what Engelking calls an "open mapping" on p. 31) iff $f$ is continuous and $f[\operatorname{int}(A)] \subseteq \operatorname{int}(f[A])$ for all subsets $A$ of $X$.

He does that latter inclusion only fact because (as opposed to continuity and closures, see 1.4.1.v ), there is no easy continuity characterisation in those "forward image plus interiors" terms. For inverse images there is 1.4.1.vi : $f$ is continuous iff

$$\forall B \subseteq Y f^{-1}[\operatorname{int}(B)] \subseteq \operatorname{int}(f^{-1}[B]).$$

and there the reverse inclusion does characterise openness (exercise 1.4.C).

1.4.18 in Engelking (what you call Theorem 1) is OK but you have to remember that "the mapping $f$ is closed" means it's continuous and closed, and ditto for open (I refer to p31 bottom yet again).

We assume throughout that $f$ is 1-1 and onto. The theorem only applies to bijections.

We need the continuity of $f$ to go from 2 or 3 back to 1. The rest is completely straightforward.

The characterisation of openness and closedness of maps are compatible between Dugundji and Engelking (it's 1.4.12 there). But of course Engelking has to assume $f$ is continuous at the start (or he cannot talk about $f$ being open or closed), but he does not use in in the proof...

Dugundji does not require open maps to be continuous so he just can state and prove them without it. The stated characterisations just hold for functions that are open iff they send open sets to open sets and ditto for closed; no continuity needed. Engelking just states it because he's chosen to define things that way.

E.g. he also defines compactness to include Hausdorffness but some of his proofs don't use that and some do. IMHO that confuses things when comparing statements with similar theorems in other textbooks sometimes.

Henno, after rereading theorem 1 in Engelking, is the function assumed to be bijective, instead of merely one to one. I know in some texts, they used "onto" in place of "to". I just want to make sure I am reading it correctly. — Seth, Sep 15 '19 at 00:23
@SethMai Yes, the function is bijective. "onto" in the context of mappings means surjective (in all the texts I know at least). If $f$ is "onto $Y$" it always that $f$ is surjective.. — Henno Brandsma, Sep 15 '19 at 05:28

Questions concerning properties of open and closed maps in general topological spaces

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