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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?

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  • $\begingroup$ An open map needn't be continuous: math.stackexchange.com/a/75601/403337 $\endgroup$ – Chris Custer Sep 12 at 3:04
  • $\begingroup$ @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. $\endgroup$ – Seth Mai Sep 12 at 3:06
  • $\begingroup$ @SujitBhattacharyya thank you for telling me that. $\endgroup$ – Seth Mai Sep 12 at 3:07
  • $\begingroup$ A “mapping” often is a term for “continuous function” and that’s the case in your first theorems. So continuity is assumed throughout. $\endgroup$ – Henno Brandsma Sep 12 at 8:38
  • $\begingroup$ @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? $\endgroup$ – Seth Mai Sep 12 at 8:48
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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.

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  • $\begingroup$ 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. $\endgroup$ – Seth Mai Sep 15 at 0:23
  • $\begingroup$ @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.. $\endgroup$ – Henno Brandsma Sep 15 at 5:28
  • $\begingroup$ Thank you for confirming that for me. $\endgroup$ – Seth Mai Sep 15 at 7:31

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