# A topological embedding is essentially an inclusion map

I want to show that if $$f\colon X \rightarrow Y$$ is a topological embedding, then there is a topological space $$Y'$$ containing $$X$$ as a subspace and a homeomorphism $$h$$ from $$Y$$ to $$Y'$$ such that $$h \circ f = j$$, where $$j\colon X \rightarrow Y'$$ is the inclusion map.

Evidently I start by taking $$Y' = X \sqcup (Y \setminus f(X))$$, the disjoint union, as a set. The definition of $$h\colon Y \rightarrow Y'$$ will be the obvious one, namely, $$h(y) = f^{-1}(y)$$ if $$y \in f(X)$$, and $$h(y) = y$$ if $$y \in Y \setminus f(X)$$. Certainly $$h$$ is a bijection satisfying $$h \circ f = j$$.

My question: How exactly should one define the topology on $$Y'$$? (I do not think it is the usual topology on the disjoint union of two topological spaces.)

Of course, a subset $$W$$ of $$Y'$$ will have to be open in $$Y'$$ if and only if $$h^{-1}(W)$$ is open in $$Y$$. So the question is how to formulate that condition in terms allowing one to verify that the inclusion map $$j\colon X \rightarrow Y'$$ is an embedding for the topology given to $$Y'$$.

(This question is of no depth whatsoever, just a matter of some fussy details!)

Note of interest: I asked the question of three different LLM chat engines and got answers that may or may not be correct but are somewhat different from one another!

• Never trust anything a LLM writes. This is especially true when it comes to math, where LLMs almost never fail to be wrong. Commented Aug 5 at 14:40
• @AlexKruckman: Actually, I've found a couple of times they gave correct proofs. (If a LLM has been fed enough of the literature, it's bound to stumble upon something correct from time to time!) But this is getting away from my question! Commented Aug 5 at 14:42
• A broken clock is right twice a day. Do not use LLMs for mathematics. Commented Aug 5 at 14:47
• Please let's get back on-topic! (I have no illusions about LLMs here.) Commented Aug 5 at 14:55
• @AlexKruckman: Yep, sorry for those typos. Commented Aug 5 at 15:07

Your setup for the question already gives you the "obvious" bijection $$h : Y \to Y'$$.

The desired topology on $$Y'$$ is characterized uniquely by the requirement that $$h$$ be a homeomorphism. In other words, a subset $$V' \subset Y'$$ is open in $$Y'$$ if and only if $$V = h^{-1}(V') \subset Y$$ is open in $$Y$$.

One thing to keep in mind here: your sentence "Of course, a subset ..." is the definition of the topology on $$Y'$$. The definition of the topology on $$Y'$$ is not "formulated" in any other terms.

But there is still a job left to do: use the definition of the topology on $$Y'$$ to prove the statement that the inclusion map $$j : X \to Y'$$ is an embedding. What that statement means is that the range restriction map $$j : X \to X$$ is a homeomorphism with respect to the subspace topology on $$X$$ that is induced from the topology on $$Y'$$.

Similarly, knowing already that the map $$f : X \to Y$$ is an embedding, what that means is that the range restriction map $$f : X \to f(X)$$ is a homeomorphism with respect to the subspace topology on $$f(X)$$ induced from $$Y$$.

So, putting this all together, given any $$U \subset X$$ we must verify that $$U$$ is open in $$X$$ with respect to the subspace topology induced from $$Y'$$ if and only if $$f(U)$$ is open in $$f(X)$$ with respect to the the subspace topology on $$f(X)$$ induced from $$Y$$.

In other words, we must prove that existence of an open $$V' \subset Y'$$ such that $$U=V' \cap X$$ is equivalent to existence of an open $$V \subset Y$$ such that $$f(U)=V \cap f(X)$$.

And that equivalence is proved by applying the fact that $$h : Y \to Y'$$ is a homeomorphism restricting to the bijection $$f : X \to f(X)$$:

• If an open $$V' \subset Y'$$ exists such that $$U=V' \cap X$$, let $$V=h^{-1}(V')$$;
• If an open $$V \subset U$$ exists such that $$f(U)=V \cap f(X)$$, let $$V'=h(V)$$.
• Yes, but how to formulate the condition on the topology so as to be able to verify that the inclusion $j$ is an embedding? That's the only thing giving me trouble. There's some annoying little detail that is eluding me. Commented Aug 5 at 15:04
• Alright, I took you a bit too literally when you wrote My question. I'll add some details. Commented Aug 5 at 15:23
• That says it, itseems to me, in as simple a way as is possible and reasonsble. Commented Aug 5 at 18:01

The condition "$$W\subseteq Y'$$ is open in $$Y'$$ if and only if $$h^{-1}(W)$$ is open in $$Y$$" is sufficient to prove directly that $$j$$ is a topological embedding.

Let $$U\subseteq X$$ be open in $$X$$. Since $$f$$ is a topological embedding, $$f(U)$$ is open in the subspace topology on $$f(X)\subseteq Y$$. So there is some $$V\subseteq Y$$ open such that $$V\cap f(X) = f(U)$$. Let $$V' = h(V)\subseteq Y'$$. Since $$h^{-1}(V') = V$$ is open in $$Y$$, $$V'$$ is open in $$Y'$$, and $$V'\cap X = h(V)\cap h(f(X)) = h(V\cap f(X)) = h(f(U)) = U.$$ So $$U$$ is open in the subspace topology on $$X\subseteq Y'$$.

Conversely, suppose $$U\subseteq X$$ is open in the subspace topology on $$X\subseteq Y'$$. Then there is some $$V'\subseteq Y'$$ such that $$V'\cap X = U$$. Let $$V = h^{-1}(V')$$, which is open in $$Y$$ by definition of the topology on $$Y'$$. We have $$V\cap f(X) = h^{-1}(h(V\cap f(X)) = h^{-1}(V'\cap h(f(X)) = h^{-1}(V'\cap X) = h^{-1}(U) = h^{-1}(h(f(U)) = f(U).$$ So $$f(U)$$ is open in the subspace topology on $$f(X)\subseteq Y$$. Since $$f$$ is a topological embedding, $$U$$ is open in $$X$$.

Of course, you can unpack this condition: For any set $$V\subseteq Y'$$, we can uniquely decompose $$V$$ as $$V = V'\cup V''$$, where $$V'\subseteq X$$ and $$V''\subseteq Y\setminus f(X)$$. Then $$V$$ is open in $$Y'$$ if and only if $$f(V')\cup V''$$ is open in $$Y$$. But it's not clear that this description would make any arguments easier.

• I'm stll being "dense" and somehow missing the detail here: Why is $V' \cap X = h(V) \cap h(f(X))$? And in the converse part why is $f^{-1}(V')$ open in $Y$? That's exactl why I asked how to reformulate (in such a way as not to mention $h$) the condition $h^{-1}(W)$ be open in $Y$. I think my problem is not being able to say in a nice way what $h^{-1}(S)$ is for subssets $S$ of $Y'$. Commented Aug 5 at 15:18
• (1) $h(V) = h(h^{-1}(V')) = V'$, and $X = j(X) = h(f(X))$, so $V'\cap X = h(V)\cap h(f(X))$. (2) My $f^{-1}(V')$ was a typo for $h^{-1}$. Fixed. (3) I added a reformulation not involving $h$ to my answer. @murray Commented Aug 5 at 15:26
• that "unpacked" condition amounts to saying that $V$ is open in $Y'$ if and only if $f(V \cap X) \cup \bigl(V \cap (Y \setminus f(X)\bigr)$ is open in $Y$, right? Commented Aug 5 at 15:33
• @murray Correct. Commented Aug 5 at 15:34
• Ok, think I got it. I don't understand wny it this simple thing was giving me trouble nailing down the detalls. It's the sort of thing that is "obvioous" or "trivial"! But then trying to write out the details produces snags. Commented Aug 5 at 15:39

What we have to do is to construct a set $$Y'$$ containing $$X$$ as a genuine subset and a bijection $$h : Y' \to Y$$ such that $$h \mid_X = f$$.

Once we have done this, we give $$Y'$$ the unique topology such that $$h$$ becomes a homeomorphism.

We claim that the subspace topology induced from $$Y'$$ on the subset $$X \subset Y'$$ is the original topology of $$X$$.

Let $$X'$$ denote the subspace of $$Y'$$ with underlying set $$X$$. Since $$h$$ is a homeomorphism, it restricts to a homeomorpism $$h ' : X' \to h(X) = f(X)$$. Since $$f : X \to Y$$ is an embedding, its codomain restriction $$f ': X \to f(X)$$ is a homeomorphism. Thus $$\phi = (f')^{-1} \circ h' : X' \to X$$ is a homeomorphism. By construction, $$\phi$$ is the identity on the level of sets. This means that $$X' = X$$ as toplogical spaces.

The problem is to find a set $$Y'$$ as above. Taking the disjoint union is a nice idea, but the standard construction of $$X \sqcup (Y \setminus f(X))$$ does not produce a set $$Y'$$ containing $$X$$ as a genuine subset; it only contains a "canonical copy" of $$X$$.

Actually we have to find a set $$Z$$ which is disjoint from $$X$$ and a bijection $$b : Z \to Y \setminus f(X)$$. This is not trivial. However, based on this, we may take $$Y' = X \cup Z$$ and $$h : Y' \to Y, h(\xi) = f(\xi)$$ for $$\xi \in X$$ and $$h(\xi) = b(\xi)$$ for $$\xi \in Z$$.

Here is a suggestion based on a bit cardinal arithmetic. Let $$Z'$$ be the power set of $$Y$$. Its cardinality is bigger than that of $$Y$$. Consider the sets $$Z_1= Z' \setminus X$$ and $$Z_2 = Z' \cap X$$. The cardinality of $$Z_2$$ is at most that of $$X$$ and thus at most that of $$Y$$ (because $$X$$ has the same cardinality as $$f(X) \subset Y$$). Hence the cardinality of $$Z_1$$ must be bigger than that of $$Y$$ (we only have to know that it has at least the cardinality of $$Y$$). We conclude that there exists an injection $$i : Y \to Z_2$$, hence an injection $$i' : Y \setminus f(X) \to Z_1$$. Now take $$Z = i'(Y \setminus f(X))$$ and $$b(z) = (i')^{-1}(z)$$.