Lifting a convergent net through a quotient map

Throughout, $\pi : X \to Y$ will denote a topological quotient map. It is sometimes desirable to understand the topology of $Y$ in terms of nets. At first, one might guess the following holds

Claim 1: If $y_i \to y$ in $Y$, then we can write $y_i = \pi(x_i)$, $y = \pi(x)$ such that $x_i \to x$ in $X$.

but easy examples show this is false, even if $X$ and $Y$ are very simple spaces.

Example 1: Let $X = [0,1]$ and $Y=S^1$ with $\pi$ the map which glues the endpoints together. Let $(a_n)$ and $(b_n)$ be sequences in $(0,1)$ converging to $0$ and $1$ respectively. Then, $\pi(a_1), \pi(b_1), \pi(a_2), \pi(b_2), \pi(a_3), \ldots$ is a convergent sequence in $Y$ whose unique lift to $X$ does not converge.

In the above example, $a_1,b_1,a_2,b_2,a_3, \ldots$ does, however, have convergent subsequences. So, one might guess the following is true.

Claim 2: If $y_i \to y$ in $Y$, then, for some subnet $y_{i_k}$ of $y_i$, we can write $y_{i_k} =\pi(x_k)$, $y=\pi(x)$ such that $x_k \to x$ in $X$.

This claim is also false, though it took me a while to think of a counterexample.

Example 2: Let $X = \mathbb{R}$, in the standard topology. Define an equivalence relation $\sim$ on $X$ such that:

• The integers $\mathbb{Z}$ are an equivalence class.
• Each open interval $(n,n+1)$, $n \in \mathbb{Z}$ is an equivalence class.

Let $Y = X / \sim$ in the quotient topology. The topology on $Y$ is as follows:

• Each $y_n = (n,n+1)$ is an open point of $Y$.
• $y_\infty = \mathbb{Z}$ has all of $Y$ as its only open neighbourhood.

Then, $y_1,y_2,y_3,\ldots$ converges to $y_\infty$ in $Y$, but cannot be lifted to a convergent sequence in $X$.

I am unsatisfied with this example, however, because $Y$ is a not even a Hausdorff space.

Question: Can we violate Claim 2 using nicer spaces $X$ and $Y$? Can they be Hausdorff? Compact Hausdorff? Compact metrizeable?

In the positive direction, we do have that Claim 2 holds when the quotient map is open.

Proposition: Suppose the quotient map $\pi : X \to Y$ is open. Let $y_i \to y$ in $Y$. Then, for any lift $x$ of $y$ there is a subnet $y_{i_k}$ of $y_i$ and lifts $x_k$ of $y_{i_k}$ such that such that $x_k \to x$.

Proof: Let $\mathscr{U}_x$ be the collection of all open sets $U \subseteq X$ with $x \in U$. Let $K = I \times \mathscr{U}_x$, where $I$ is the index set of $y_i$. Define $k \mapsto i_k : K \to I$ and $k \mapsto x_k : K \to X$ as follows.

For $k=(i,u)$:

• Choose $i_k$ so that $i_k \geq i$ and $y_{i_k} \in \pi(U)$.
• Choose $x_k$ so that $x_k \in U$ and $\pi(x_k) = y_{i_k}$.

It is simple to see that $y_{i_k}$ is a subnet of $y_i$ and $x_k \to x$.

So, in any answer to my question, $\pi : X \to Y$ cannot be an open map.

This is not possible if $X$ is compact and $Y$ is Hausdorff. In that case, if $(y_i)$ converges to $y$ in $Y$, pick any elements $x_i\in X$ such that $\pi(x_i)=y_i$. Since $X$ is compact, some subnet of $(x_i)$ converges to some point $x\in X$. Since $\pi$ is continuous, the image of this subnet must converge to $\pi(x)$. But the image of the subnet is a subnet of $(y_i)$, and so converges to $y$. Since $Y$ is Hausdorff, this implies $y=\pi(x)$.

On the other hand, it is possible with Hausdorff spaces. Indeed, you can just modify your example slightly: let $X=\mathbb{R}$, and take the quotient $Y$ in which $\mathbb{Z}$ is a single equivalence class but all the other equivalence classes are singletons. It is easy to verify that this quotient $Y$ is Hausdorff. However, you can find a net $(x_i)$ of non-integers in $X$ which goes to infinity such that $\pi_i(x_i)$ converges to $\mathbb{Z}$ in $Y$ (just take any net associated to the filter of sets which are a deleted neighborhood of all but finitely many integers). Such a net has no convergent subnets in $X$, so this violates Claim 2.