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I think something is missing in the definition of homeomorphism I saw. It just said it maps the collection of open sets to the collection of open sets in a bijective way. What exactly makes this preserve topology? I can think of weird situations where each individual open set is mapped to totally disjoint open sets in a bijective way, throwing the topological structure out the window.

J. W. Tanner
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Emil
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  • What's your question? – Jakobian Jun 21 '23 at 18:07
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    Perhaps what you heard was a casual description of a basic property of a homeomorphism to help develop intuition. The actual definition is stronger. A homeomorphism $f: X \to Y$ is a bijective map that is continuous and its inverse is also continuous. In other words, the nested structure of open neighborhoods of a given point is also preserved. – Sammy Black Jun 21 '23 at 18:08
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    Open sets are the topological structure – J. W. Tanner Jun 21 '23 at 18:11
  • Ok, so it must be the definition of continuous I saw that was wrong. Does it need to map open sets and all open sets in the neighbourhood to open sets related in the same way in the codomain's neighborhood somehow? I read it as only requring each open set to be mapped to an open set. – Emil Jun 21 '23 at 18:13
  • No, topological structure also includes some kind of order relation between the open sets, as induced by the axioms of a topological space. An open set itself is just some set you think is open. – Emil Jun 21 '23 at 18:16
  • The order relation on topology $\tau$ on $X$ is already included as inherited from the partial order of all subsets of $X$. There's no need to mention it, we're not doing pointless topology. – Jakobian Jun 21 '23 at 18:32
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    (+1) I think this is an interesting question for beginners that would be useful to have an archived MSE answer, but the original wording (perhaps due to ESL issues?) doesn't quite ask what I believe was intended (based on earlier comments), so I've revised the question to reflect what I believe was intended. – Dave L. Renfro Jun 21 '23 at 18:32
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    @DaveL.Renfro that's not true, questions like this already appeared on this site. See here for example. There's no need to "archive" it. – Jakobian Jun 21 '23 at 18:39
  • @Jakobian: there’s no answer with that question – J. W. Tanner Jun 21 '23 at 18:50
  • @J.W.Tanner but that question has good description of the problem. On the other hand, this question was edited and who knows what the author originally meant. – Jakobian Jun 21 '23 at 18:52
  • Yeah it seems to be that map is used in a different way from what I am used to. I have no idea why I did not notice this, I have read tonnes of math books, maybe it is because I gloss over the text and look at the symbols :') I am used to saying a map from X to Y, but that it maps element x to element y. So the issue is there somewhere. – Emil Jun 21 '23 at 18:59
  • Maybe what the OP is thinking is: the definition brings to mind a bijective set map between the corresponding frames which for example wouldn't necessarily have to preserve the partial orders - but in fact, the bijection of frames is induced by a function of the point sets $f : X \to Y$, then the map between frames has to be automatically a morphism of frames (so it respects order in particular). (To OP: "frame" is a fancy term for an "algebraic" structure on the collection of open sets, which includes a partial order, a binary intersection operation, and an infinitary union operation.) – Daniel Schepler Jun 21 '23 at 19:01
  • Can you please quote or cite the actual definition that you are asking about, rather than just give a loose paraphrase. Not being precise about a definition you are querying is a frequent cause of wasted time on MSE. – Rob Arthan Jun 21 '23 at 22:04
  • See e.g. the first sentence in https://en.m.wikipedia.org/wiki/Open_and_closed_maps. It says it maps open sets to open sets. For some weird reason that actually means the function's image and preimage is open, and not that it maps individual open sets to open sets (i.e. the elements being mapped are not the open sets themselves). Seems to be some kind of oddity in mathematical english that I am not comfortable with. – Emil Jun 21 '23 at 22:27
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    I think you have correctly identified the source of your confusion: "function between two topological spaces" is being used to mean "function between the underlying sets of two topological spaces" and the phrase "maps open sets to open sets" is about the images of open sets under the function on the underlying sets. The notation $f(U)$ for the image of a set $U$ under a function on the points of $U$ is indeed confusing, but, unfortunately, standard. – Rob Arthan Jun 21 '23 at 22:44
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    As an alternative, another notation for the image I started using and enjoy, is $f[U]$. – Jakobian Jun 22 '23 at 01:18
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    @Jakobian: me too - I learnt it from Bob Solovay. Let's promote $f[U]$. – Rob Arthan Jun 22 '23 at 22:18

2 Answers2

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The definition you heard is almost correct if put precisely, albeit non-standard. Let $X$ and $Y$ be topological spaces, and $f: X \to Y$ a map between them. Then $f$ is a homeomorphism if the image map from $\mathcal{P}(X)$ to $\mathcal{P}(Y)$ defined by $$ U \mapsto f(U) = \{ f(x) | x \in U \} $$ gives a bijection between the open sets of $X$ and the open sets of $Y$, and $f$ is a bijection$^{\ast}$.

Does this preserve the order structure between sets? Yes, it does, even if $f$ is not a homeomorphism. That is, if $U_{1} \subseteq U_{2}$ are subsets of $X$, then necessarily $f(U_{1}) \subseteq f(U_{2})$. That is, $$ f(U_{1}) = \{ f(x) | x \in U_{1} \} \subseteq \{ f(x) | x \in U_{2} \} = f(U_{2}). $$

In other words, so long as $f$ is defined as a map between $X$ and $Y$, it is true. If I understand correctly, it seems like your confusion comes from assuming that only the bijection between the collections of open sets is given, and not the underlying map $f$.

Alternatively, the standard definition of a homeomorphism from $X$ to $Y$ is a map $f: X \to Y$ which is continuous with continuous inverse; $f$ is continuous if the preimage map $$ V \mapsto f^{-1}(V) = \{ x \in X | f(x) \in V \} $$ maps open sets to open sets. It's a good exercise to prove that these two definitions are equivalent. The key takeaway in my mind is to understand how the properties of a continuous function are contained both in the function itself and its induced maps (image map, pre-image map).

$\ast$: as pointed out by Daniel Schepler, it is essential to add that $f$ is a bijection. For example, the constant map from $\{1, 2\}$ to $\{ 1 \}$ cannot be a homeomorphism yet is a bijection between open sets when both have the indiscrete topology (only the empty set and the whole space open).

Jacob Maibach
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  • I don't understand why the subset relation is preserved? Can't $U_1$ and $U_2$ be mapped to disjoint, but open, sets in Y? – Emil Jun 21 '23 at 18:49
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    Okay, something is starting to become clear to me now. The function does not map open sets to open sets like I think, it maps elements of open sets to elements of open sets. – Emil Jun 21 '23 at 18:55
  • @Emil I'd recommend playing around with the case where $X$ and $Y$ are small finite sets (say ${1, 2, 3}$ and ${a, b, c, d}$) to get intuition for the image map. It's quite useful to know its basic properties front and back when studying topology. – Jacob Maibach Jun 21 '23 at 19:01
  • @Emil The function definitely maps open sets to open sets – Alex Mathers Jun 21 '23 at 19:02
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    I thought it was a set to set function, but continuity is about the image and preimage, a bit embarrassing I forgot about that. – Emil Jun 21 '23 at 19:02
  • Like in your example, I thought $x={1,2,3}$ and $f(x)={a,b,c,d}$. – Emil Jun 21 '23 at 19:04
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    Just inducing a bijection on the collections of open sets is not enough to get a homeomorphism; the function also needs to be a bijection itself. (Counterexample: take ${ 1, 2 }$ with the indiscrete topology and ${ 1 }$ with the unique topology; then the unique function ${ 1, 2 } \to { 1 }$ induces a bijection on the collections of open sets. Though fans of pointless topology would doubtless argue that this is close enough to a homeomorphism to still be interesting...) – Daniel Schepler Jun 21 '23 at 19:25
  • The "right" definition of a homeomorphism is a continuous bijection with a continuous inverse. As @DanielSchepler points out your proposed alternative doesn't guarantee that $f$ is a bijection. I'm intrigued to know whether your condition does guarantee that $f$ is continuous: I suspect not, but haven't found a counter-example yet. – Rob Arthan Jun 21 '23 at 22:34
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Do you agree that knowing which subsets are open subsets is the same as knowing the topology? Then prove this lemma using the definition of homeomorphism and you should feel content:

Lemma: Suppose $f:X\to Y$ is a homeomorphism. Then a subset $U$ of $X$ is open if and only if the subset $f(U)$ is an open subset of $Y$. Also (and equivalently) a subset $V$ of $Y$ is open if and only if $f^{-1}(V)$ is an open subset of $X$.

You can also go through and take any topological property you like and show that if $X$ and $Y$ are homeomorphic then $X$ has a property $P$ if and only if $Y$ has property $P$, for instance with $P\in\{\text{metrizable,hausdorff,compact,connected,path-connected}\}$

Alex Mathers
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