Quotient Spaces (from Viro, Harlamov et al)

Question

dears! I have some troubles with quotient spaces.

1. Let $A\subset X$ and $X=\{b_1,b_2, \dots, b_n, \dots, a_1,\dots, a_n,\dots\}$ $A=\{a_1,a_2,\dots,a_n,\dots\}$. Is it correct to imagine, think about quotient space $X/A$ like space, where if $x \not\in A$ then x is "independent" element $x$ (e.g. $b_1$), but if $x\in A$ then we get new point $\alpha$, where $\alpha = a_1 \sim a_2\sim\dots$?

2. Typical exercise. Let $f \colon X \to Y$ be a homeomorphism and $A \subset X$. Show that $X/A$ and $Y/f(A)$ is homeomorphic.

My idea: $\hat{f}$ is a homeomorphism from $X/A$ to $Y/f(A)$ and $\hat{f}(x)=f(x)$.

Step 1: If (1.) is correct then X/A is $\{b_1,\dots,b_n,\dots, \alpha \}$ and $x \not\in A \Rightarrow$ $f$ continuous in $x$ and we can use results for $f$ (e.g. take some neighborhoods and check the conditions).

Step 2: If $x\in A \Rightarrow$ in $X/A:$ $\alpha=a_1\sim a_2\sim\dots$ and we can define $\beta=\hat{f}(\alpha)\sim f(a_1)\sim f(a_2)\sim\dots$ and:

i) We can define neighborhood of $\alpha$ as union of neighborhoods $U_{a_1}, U_{a_2},\dots, U_{a_l},\dots$?

ii) We can say that f is continuous at $a_1 \in X$ and we can define other equivalent points as "a_1" and use properties $f$?

iii) If my ideas isn't correct please help me to understand.

Thanks.

Answer 1

Your idea is good.

Let $X \sim Y$ with $f: X \to Y$ a homeomorphism; let $A \subseteq X$ be an arbitrary set in $X$. Let $X / A$ and $Y / f [A]$ be the quotient spaces obtained by identifying all points in $A$ in $X$ and all points $f [A]$ in $Y$, respectively, and let $p_X: X \to X / A$ and $p_Y: Y \to Y / f [A]$ be the respective quotient maps.

We want to show that $X / A \sim Y / f [A]$. Consider $\hat{f}: X / A \to Y / f [A]$ given by $\hat{f} (B) = f [B]$. Surjectivity of $\hat{f}$ follows directly from the surjectivity of $f$. Injectivity of $\hat{f}$ follows from the fact that fact that given $B \ne C$ in $X / A$, the preimages $p_X^{-1} (C), p_X^{-1} (D)$ are disjoint in $X$ along with the injectivity of $f$. Therefore, $\hat{f}$ is bijective, and $\hat{f}^{-1}$ is well-defined.

To show that $\hat{f}$ is continuous, it is sufficient to show that $\hat{f} \circ p_X: X \to Y / f [A]$ is continuous (I'll leave the rather straightforward proof of this property of quotient maps to you). So, let $V$ be open in $Y / f [A]$. By continuity of the composition $p_Y \circ f: X \to Y / f [A]$, we have that $(p_Y \circ f)^{-1} [V]$ is open in $X$. Note that $\hat{f} \circ p_X = p_Y \circ f$ (show!), so that $(\hat{f} \circ p_X)^{-1} [V] = (p_Y \circ f)^{-1} [V]$. This shows $(\hat{f} \circ p_X)^{-1} [V]$ is open in $X$, as desired.

Showing that $\hat{f}^{-1}$ is continuous is basically the same: it is sufficient to show that $\hat{f}^{-1} \circ p_Y: Y \to X / A$ is continuous. Let $U$ be open in $X / A$; use continuity of the composition $p_X \circ f^{-1}$ to conclude that $(p_X \circ f^{-1})^{-1} [U]$ is open in $Y$; then, since $p_X \circ f^{-1} = \hat{f}^{-1} \circ p_Y$, we have that $(f^{-1} \circ p_Y)^{-1} [U]$ is open in $Y$, as desired.