Embedding of a quotient topological space

Question

Recently, I solved the following problem:

Let $B, X$ be topological spaces, $A\subset B$ a subset and $i_B:B\to X$ an embedding. Then there exists an embedding $i_N:B/A\to X/A$.

For this exercise we can consider the commutative diagram

$$\require{AMScd} \begin{CD} B @>{i_B}>> X\\ @V{\pi_B}VV @V{\pi_X}VV \\ B/A @>{i_N}>> X/i_B(A)\end{CD}$$

where $\pi_B$ and $\pi_X$ are the respective canonical maps.

So, having defined $i_N$ as $i_N(\pi_B(b))=\pi_X(i_B(b))$, the important thing is basically to show $i_N$ is open/closed with respect to $i_N(B/A)$, for which we use $A$ is open/closed:

If $U\subset B/A$ is open/closed, then $\pi_B^{-1}(U)$ is open/closed, so $i_B(\pi_B^{-1}(U))$ is open/closed with respect to $i_B(U)$, so there exists $W\subset X$ open/closed such that $i_B(\pi_B^{-1}(U))=W\cap i_B(B)$. Note that $\pi_X^{-1}(\pi_X(W))=W$, since either $i_B(A)\subset W$ or $W\cap i_B(A)=\emptyset$ (depending on whether $\pi_B(A)\in U$ or not). Hence, no matter the case, $\pi_X(W)$ is open/closed in $X/i_B(A)$. Knowing that $$\begin{align}i_N(U)&=\pi_X(i_B(\pi_B^{-1}(U)))=\pi_X(W\cap i_B(B))=\pi_X(W)\cap\pi_X(i_B(B))=\\&=\pi_X(W)\cap i_N(B/A)\end{align}$$ (the second last equality is true for the same reason we exposed before), we conclude the proof.

Now, I have a silly question and a serious one:

1. Out of curiosity, does anybody know why they named the embedding $i_N$? (Please, write a comment if you're answering just this one question).

2. Is there a generalization of this result?

Any help will be appreciated. Thanks!

Answer 1

After a lot of time thinking about it, I think I finally had a breakthrough.

In general, as an answer for 2., we have the following result:

Let $B,X$ be topological spaces, $i_B:B\to X$ an embedding and $\mathcal {R}_1, \mathcal R_2$ equivalence relations (in $B$ and $X$ respectively) such that $x\overset{\mathcal R_2}\sim y$ iff $\exists\, b,c\in B:i_B(b)=x,i_B(c)=y,b\overset{\mathcal R_1}\sim c$. Then there exists an embedding $i_N:B/\mathcal R_1\to X/\mathcal R_2$.

The condition basically means that $\mathcal R_2$ must preserve via $i_B$ the relations from $\mathcal R_1$ in $B$ (that's the if direction); and that it must not create new relations in $X$ that aren't in $i_B(B)$ (or, in other words, every element in $X\setminus i_B(B)$ must be alone in its equivalence class).

Proof:

Let $\pi_B:B\to B/\mathcal R_1$ and $\pi_X:X\to X/\mathcal R_2$ be the canonical projections. The proof has three parts:

• Definition and bijectivity:

Consider $d=\pi_B(b)\in B$ and define $i_N(d)=i_N(\pi_B(b))=\pi_X(i_B(b))$ (note we have $i_N\circ \pi_B=\pi_X\circ i_B$). It is well-defined, since $b\overset{\mathcal R_1}\sim c\Rightarrow i_B(b)\overset{\mathcal R_2}\sim i_B(c)$, and it's clearly surjective since $i_N(\pi_B(B))=\pi_X(i_B(B))=i_B(B)/\mathcal R_2$.

If $i_N(\pi_B(b))=i_N(\pi_B(c))$ then $\pi_X(i_B(b))=\pi_X(i_B(c))$, which means $i_B(b)\overset{\mathcal R_2}\sim i_B(c)\Rightarrow b\overset{\mathcal R_1}\sim c\Rightarrow \pi_B(b)=\pi_B(c)$, so $i_N$ is injective.

• Continuity:

Consider an open subset $V\subset X/\mathcal R_2$, then, by definition, $i_N^{-1}(V)$ is open iff $\pi_B^{-1}(i_N^{-1}(V))$ is open, but $\pi_B^{-1}(i_N^{-1}(V))=i_B^{-1}(\pi_X^{-1}(V))$, which is open by hypothesis.

• $i_N$ is open/closed respect to its image:

We can prove $i_N=\pi_X\circ i_B\circ \pi_B^{-1}$: Take $d=\pi_B(B)\in B/\mathcal R_1$. Hence, in consequence of the condition (the if direction), $i_N(d)=i_N(\pi_B(b))=\pi_X(i_B(b))=\pi_X(i_B(\pi_B^{-1}(\pi_B(b)))=\pi_X(i_B(\pi_B^{-1}(d)))$.

Now let's consider $A\subset B/\mathcal R_2$ an open/closed subset. Then $i_B(\pi_B^{-1}(A))$ is open/closed in $i_B(B)$, which means $i_B(\pi_B^{-1}(A))=i_B(B)\cap W$ with $W\subset X$ open/closed. We'll prove $\pi_X^{-1}(\pi_X(W))=W$:

Trivially, $W\subset \pi_X^{-1}(\pi_X(W))$. On the other hand, if $w\in \pi_X^{-1}(\pi_X(W))$ and $w\notin W$ then there exists $y\neq w:y\in W,\pi_X(y)=\pi_X(w)$. Hence $w\overset{\mathcal R_2}\sim y$ and therefore $\exists\, b,c\in B:i_B(b)=w,i_B(c)=y, b\overset{\mathcal R_1}\sim c$. Then $y\in i_B(B)$, so $y\in i_B(\pi_B^{-1}(A))$, which means (since $i_B$ is injective) that $c\in \pi_B^{-1}(A)$. Knowing that $b\overset{\mathcal R_1}\sim c$ we have that $\pi_B(b)=\pi_B(c)\in\pi_B(A)\Rightarrow b\in\pi_B^{-1}(A)$, so $w=i_B(b)\in i_B(\pi_B^{-1}(A))$, which is a contradiction. Therefore $w\in W$ necessarily. So we have that $\pi_X^{-1}(\pi_X(W))=W$ and thus $\pi_X(W)$ is open/closed in $X/\mathcal R_2$.

Finally,

$\begin{align}i_N(A)&=\pi_X(i_B(\pi_B^{-1}(A)))=\pi_X(i_B(B)\cap W)=\pi_X\Big(i_B\cap \pi_X^{-1}(\pi_X(W))\Big)\\&=\pi_X(i_B(B))\cap\pi_X\Big(\pi_X^{-1}(\pi_X(W))\Big)=i_B(B)/\mathcal R_2\cap\pi_X(W),\end{align}$

which is open/closed.

This concludes the proof.