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?
Quotient map from $\mathbb R^2$ to $\mathbb R$ with cofinite topology
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general-topology
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1What topology does $\mathbb{R}^2$ have? The standard one? – 2012-02-19
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0Yeah, standard topology on $\mathbb R^2$. – 2012-02-19
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0What do you precisely mean by this (maybe it's just me who doesn't understand)? – 2012-02-21
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0A 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. – 2012-02-25