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Let $X = \mathbb R$ under the cofinite topology. Is there a quotient map $q : \mathbb R^2 \rightarrow X$? Intuitively, this seems like it should be false, since $\mathbb R^2$ has "too many" open sets. However, I am not sure how to prove it. Any ideas?

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What topology does $\mathbb{R}^2$ have? The standard one? – Chris Eagle Feb 19 '12 at 22:10
Yeah, standard topology on $\mathbb R^2$. – user15464 Feb 19 '12 at 22:56
What do you precisely mean by this (maybe it's just me who doesn't understand)? – Matt Feb 21 '12 at 17:48
A quotient map is a continuous surjective map $q : X \rightarrow Y$ such that $U\subset Y$ is open if and only if $q^{pre}(U)$ is open in $X$. I am asking whether such a map exists in the case where $X = \mathbb R^2$ (usual topology) and $Y = \mathbb R$ with the cofinite topology. – user15464 Feb 25 '12 at 18:27

There is no such $q$.

(Edit: As pointed out in the comments, what this proves is that there is no continuous open surjection from $\mathbb{R}^2$ to $X$. Of course, a continuous open surjection is a quotient map, but there are quotient maps which are not open. So, this doesn't answer the problem in the generality its asked.)

To see this, we begin with a characterization of continuous surjective maps $q:\mathbb{R}^2\rightarrow X$.

Lemma 1: The data of $q$ is equivalent to a partition of $\mathbb{R}^2$ into $\mathfrak{c} = |\mathbb{R}|$ pairwise disjoint closed sets together with a choice of bijection from the partition to $\mathbb{R}$.

Proof: Given $q$, the partition of $\mathbb{R}^2$ is given by $F_p = q^{-1}(p)$ as $p$ ranges over $X$. Each $F_p$ is closed since $q$ is continuous and $\{p\}$ is closed in $X$. Each $F_p$ is nonempty since $q$ is surjective. They are clearly pariwise disjoint. Finally, they cover $\mathbb{R}^2$ since $s\in \mathbb{R}^2$ is in $F_{q(s)}$. The bijection between the partition and $\mathbb{R}$ sends an $F_p$ to $p$.

Conversely, given a partition of $\mathbb{R}^2$ into pairwise disjoint closed sets and a bijection from this partition to $\mathbb{R}$, define $q(F_p) = p$. Then $q$ is surjective. It is also continuous as the only closed sets of $X$ are finite collections of points. Then $q^{-1}(\{p_1,...,p_n\}) = F_{p_1}\cup...\cup F_{p_n}$ is a finite union of closed sets, hence closed. $\square$

Now, with this characterization, we have

Lemma 2: A continuous surjective map $q$ is open iff every nonempty open subset of $\mathbb{R}^2$ intersects all but finitely many of the $F_p$.

Proof: If $q$ is open, then $q(U)$ is open for every open $U\subseteq \mathbb{R}^2$. Open sets are precisely the complement of finite sets, so $q(U) = X-\{p_1,...,p_n\}$. Then $F_p\cap U$ is nonempty unless $p \in\{p_1, p_2,..., p_n\}$.

Conversely, if $U$ intersects all but finitely many $F_p$ with $p\in\{p_1,...,p_n\}$, then $q(U) = X-\{p_1,..., p_n\}$, so $q$ is open. $\square$

Finally, we to see there no continuous open $q$, have

Lemma 3: There is no partition of $\mathbb{R}^2$ into $\mathfrak{c}$ closed sets for which every open set intersects all but finitely many of the closed sets in the partition.

Proof: We proceed by contradiction. So, assume $\{F_p\}$ is such a partition.

Given an open set $U$ with rational center and rational radius, call $F_p$ "bad with respect to $U$" if $F_p\cap U = \emptyset$. Call $F_p$ "bad" if it is bad with respect to some $U$ with rational center and rational radius.

I claim there are only countably many bad $F_p$. To see this, notice simply that there are only countably many such $U$ and for each $U$, there are only finitely many $F_p$ which are bad with respect to $U$.

Since we have $\mathfrak{c}$ closed sets in our partition, and only countably many bad ones, there must be a closed set $F$ in the partition which is not bad. This $F$ then intersects every $U$ with rational center and rational radius. Since these $U$ form a base for the topology on $\mathbb{R}^2$, this $F$ must intersect every open set. In other words, $F$ is dense. Since $F$ is also closed, it follows that $F = \mathbb{R}^2$. But $F$ was one part of a partition of $\mathbb{R}^2$ into $\mathfrak{c}$ many closed sets. This contradiction establishes lemma 3. $\square$

Putting these together, by lemma 1 and 2, finding a quotient map $q:\mathbb{R}^2\rightarrow X$ is the same as finding a partition of $\mathbb{R}^2$ into $\mathfrak{c}$ many closed sets for which every open set intersects all but finitely many of the closed sets in the partition. By lemma 3, this can't happen, so there is no quotient map.

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A similar solution was suggested in this question. – t.b. Jun 24 '12 at 18:53
I think you've shown that there's no open continuous map from $R^2$ onto $X$, not that there's no quotient map. That's the same mistake as the one I've made in the suggestion t.b.'s referring to. – tomasz Jun 24 '12 at 19:13
I think the problem is a lot harder than it looks, and might even be impossible to solve. – tomasz Jun 24 '12 at 19:17
@tomasz: I somehow forgot that not every quotient map is open. I'll keep thinking about it. What do you mean by "impossible to solve", do you think it may be independent of ZFC? – Jason DeVito Jun 24 '12 at 20:58
@JasonDeVito: yeah, that's what I usually think when I see an apparently very hard (semi)combinatorial problem. :) It's just a hunch, though. – tomasz Jun 25 '12 at 0:46

We suppose such a $q$ exists.

We define a map $D:X \times X \rightarrow \mathbb{R}$ by $D(x,y)= d(q^{-1}\{x\}, q^{-1}\{y\})$, where $d$ is the distance in $\mathbb{R}^2$.

Edit: We define $T: X \times X \rightarrow \mathbb{R}$ by $T(x,y)= \inf\{ \sum_{i=0...n-1} D(x_i, x_{i+1}) | n \in \mathbb{N}, x_0=x, x_n=y, x_i \in X\}$

We have $T(x,z) \leq T(x,y)+ T(y,z)$, for all $x,y,z \in X$.

We define a relation $S$ in $X$ by $S(x,y)$ is true if and only if $T(x,y)=0$.

The quotient space $Y:=\mathbb{R}/S$ is a metric space. The distance is $\overline{T}$.

Let $\psi: X \rightarrow Y$ the map $x \mapsto \overline{x}$.

$\psi \circ q$ is continuous, because:

$\overline{T}(\psi \circ q(M),\psi \circ q(N)=T(q(M),q(N) \leq d(q^{-1}(\{q(M)\}),q^{-1}(\{q(N)\}) \leq d(M,N)$,

for all $M,N \in \mathbb{R}^2$

So $\psi$ is continuous. Indeed, let $U$ an open set in $Y$, $q^{-1}(\psi^{-1}(U))$ is an open set in $\mathbb{R}^2$. So by definition of $q$, $\psi^{-1}(U)$ is open in $X$.

Let $K$ a compact set in $\mathbb{R}^2$. $\psi \circ q(K)$ is compact. $Y$ is Hausdorff, so $\psi \circ q(K)$ is closed.

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Why would we have $T(x,z)\leq T(x,y)+T(y,z)$? It is easy to find a counterexample for general, even general quotient maps. Just take the border of a square with vertices cut off along with small neighbourhood, then collapse what remains of some three edges. The distance between opposing edges is the length of the side of the original square, but the distance between two neighboring edges is very small. Unless you meant Hausdorff distance, for which it is not clear that it is finite. – tomasz Jun 24 '12 at 14:40
Why? I've just given a counterexample for this. – tomasz Jun 24 '12 at 14:47
Yes, you are right... I edit. – francis-jamet Jun 24 '12 at 14:56
When proving compactness, there is the case where the preimage of $\overline x_n$ is the entire $X$. So you need to say that we can choose a subsequence that is not the entire $Y$. Unless $Y$ is a singleton set. But then compactness is obvious. – tomasz Jun 24 '12 at 15:24
Yes, I don't know why I wanted to prove compactness. There is another problem: if $K$ is compact and $q(K)$ is all $X$, $q(K)$ is closed but not finite. – francis-jamet Jun 24 '12 at 16:27

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