Quotient map from $\mathbb R^2$ to $\mathbb R$ with cofinite topology 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?
 A: 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.
A: 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.
