Proving a function between quotient spaces is continuous

Question

This is a question related to a problem previously asked here and also here. It states that the infinite strip $\mathbb{R}\times\left[\frac{-1}{2},\frac{1}{2}\right]$ under the group action $m\cdot(x,y) = (m+x,(-1)^my)$ ($m\in \mathbb{Z}$) is homeomorphic to the Mobius strip (I mean, the quotient $X/\mathbb{Z}$ is homeomorphic to the Mobius strip).

Denote $X$ as the infinite strip $\mathbb{R}\times\left[\frac{-1}{2},\frac{1}{2}\right]$ and $\sim_1 $ the relationship resulting from the group action. Denote also $M$ as the unit square and $\sim_2$ as the relationship $(0,y) \sim_2 (1,1-y)$, so $M/\sim_2$ is the Mobius strip. I think the function $f: X/\sim_1 \to M/\sim_2$ given by $$ f( [x,y]_1 ) =[ (x-\lfloor{x}\rfloor,(-1)^{\lfloor{x}\rfloor}y+1/2) ]_2$$ should help proving one direction of the homeomorphism (I also saw this function as a suggestion in a solution). But I am really having a hard time proving that this function is indeed continuous.

I was thinking something like this: Consider $\pi_1 : X\to X/\sim_1$ and $\pi_2 : M\to M/\sim_2$ the quotient maps under the corresponding relations $\sim_1$ and $\sim_2$. Then, $\pi_2F = f\pi_1$, where $F:X \to M$ is a suitable continuous function. Then $f\pi_1$ is continuous (since $\pi_2$ and $F$ are) and finally $f$ is also continuous by the universal mapping property. However, I tried simply putting $F(x,y) = (x-\lfloor{x}\rfloor,(-1)^{\lfloor{x}\rfloor}y+1/2) $ but I am afraid this is not continuous (considering near points to the left and right of $(1,0)$ for example). So, is this the usual way of proving continuity between two quotient spaces? There is another approach for proving something like this? Thanks in advance!!

Answer 1

Suppose we have a topological space $X$ and a relation $\sim$ on it, with projection $\pi : X \to X / \sim$. Then for every continuous $f : X \to Y$ such that for all $x \sim y$, $f(x) = f(y)$, there is a unique continuous $g : X / \sim\to Y$ such that $g \circ \pi = f$. That is, if we have a function $g : X / \sim \to Y$, $g$ is continuous iff $g \circ \pi$ is continuous.

Let us take your specific example. You have $X_1 = \mathbb{R} \times [-1/2, 1/2]$ with equivalence relation $\sim_1$ and $X_2 = [0, 1]^2$ with equivalence relation $\sim_2$. We have defined the function

$h : X_1 \to X_2$

$h(x, y) = (x - \lfloor x\rfloor), (-1)^{\lfloor x\rfloor} y + 1/2)$

It suffices to show that $\pi_2 \circ h$ is continuous and that it respects the $\sim_1$ equivalence relation. I will focus on showing that $\pi_2 \circ h$ is continuous.

Consider an arbitrary open set $U \subseteq X_2 / \sim_2$. We wish to show that $S = (\pi_2 \circ h)^{-1}(U) = h^{-1}(\pi_2^{-1}(U))$ is open. Indeed, consider some $(x, y) \in S$. For hopefully obvious reasons, we need only worry about the case is when $x$ is an integer. WLOG, suppose $x = 0$. Since $h$ is continuous on $[0, 1) \times [-1/2, 1/2]$, we may take some $\delta_1 > 0$ (WLOG $<1$) such that $\{(a, b) \in B_{\delta_1}((x, y)) : a \in [0, 1)\} \subseteq S$.

Now consider that since $h(0, y) = (0, y + 1/2) \in \pi_2^{-1}(U)$, so is $(1, 1 - (y + 1/2)) = (1, -y + 1/2)$. We can define $g(a, b) = (a + 1, -b + 1/2)$ which coincides with $h$ over the domain $[-1, 0) \times [-1/2, 1/2]$. Since $g$ is continuous and $g(x, y) = (1, -y + 1/2) \in \pi_2^{-1}(U)$, we may choose some $\delta_2 > 0$, WLOG $\delta_2 < 1$ s.t. $B_{\delta_2}((x, y)) \subseteq g^{-1}(\pi_2^{-1}(U))$ and therefore $\{(a, b) \in B_{\delta_2}((x, y)) : a < 0\} \subseteq S$.

Then we see that taking $\delta = \min(\delta_1, \delta_2)$, we have $B_\delta(x, y) \subseteq S$.

Then $\pi_2 \circ h$ is continuous. Then $f$ is continuous.