A topological embedding is essentially an inclusion map

Question

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!

Answer 1

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 $V'=h(V)$.

Answer 2

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 $X$.

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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.

Answer 3

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 $b(z) = (i')^{-1}(z)$.