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Let $\pi_1 \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be projection on the first coordinate. Let $A$ be the subspace of $\mathbb{R} \times \mathbb{R}$ consisting of all points $x \times y$ for which either $x \geq 0$ or $y = 0$ (or both); let $q \colon A \to \mathbb{R}$ be obtained by restricting $\pi_1$.

Then $q$ is a surjective map.

How to show that $q$ is a quotient map that is neither open nor closed?

Saaqib Mahmood
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2 Answers2

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To show that $q$ is a quotient map, you should probably just show directly that $q$ sends saturated open sets to open sets. Note that the saturated open sets are precisely those open sets $U$ such that if $(x,y)\in U$ with $x \ge 0$, then $\{x\} \times \Bbb R \subset U$.

Note that projection maps in general aren't closed: If you can figure out an example to show that $\pi_1$ isn't closed, then that technique will probably work for $q$ as well. One idea might be to choose a countable, non-closed set in $\Bbb R$, and to express it as a projection of a countable subspace of $\Bbb R^2$ consisting of isolated points.

Projection maps are open (by the definition of the product topology), so you are going to need to use something special about $A$ to find a suitable counterexample. Intuitively, the place where $q$ behaves badly is the line $x = 0$, since to the left $q$ acts as the identity and to the right $q$ acts as a projection. A reasonable open subspace whose image isn't open is therefore going to need to intersect $x=0$. Try an example and see if it works!

Saaqib Mahmood
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Rolf Hoyer
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  • the first paragraph of your answer is clear, except that I'm unable to figure out how $q[U]$ would be open in $\Mathbb{R}$. – Saaqib Mahmood Apr 27 '15 at 05:57
  • in the second paragraph of your answer, one set I can think of is the set of all numbers $1/n$ for $n = 1, 2, 3, \ldots$, but would this work? This set is going to be the image under $q$ of the set ${1/n \colon n \in \mathbb{N} } \times \mathbb{R}$, which is not closed either. – Saaqib Mahmood Apr 27 '15 at 05:59
  • The set ${1/n}$ is a good example of a countable, non-closed set. Now you just need a suitable choice of coordinates $y_n$ such that ${1/n, y_n}$ is closed in $\Bbb R^2$. – Rolf Hoyer Apr 27 '15 at 06:02
  • Note that with this description of saturation gives the key property $q(U) = U \cap \Bbb R \times {0}$. – Rolf Hoyer Apr 27 '15 at 06:03
  • can you please elaborate on your own hints? I'm sorry, but I'm unable to come up with any ideas on how to get myself through. – Saaqib Mahmood May 03 '15 at 07:17
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Assuming the $\mathbb R \times \mathbb R $ and $\mathbb R$ spaces are equipped with the standard topology or equivalently the standard Euclidean metric, a very straightforward way of showing that $q$ is a quotient map could be as follows:

We indeed show that if $q^{-1}(U)$ is open in $A$, then $U$ is open in $\mathbb R.$ By contradiction, let's assume $U$ is not open. So, there is $x_0 \in U$ such that for any $r>0$ there is $y_r$ such that $y_r \in B (x_0,r)$ while $y_r \notin U$, where $B (x_0,r)$ is the standard ball centered at $x_0$ with radius $r$. Now, note that $(x_0,0) \in q^{-1}(U)$, so there must exist an (basic) open set of the form $B((x_0,0), r_0) \cap A$ such that $B((x_0,0), r_0) \cap A \subset q^{-1}(U)$, where $B((x_0,0), r_0)$ is the usual ball centered at $(x_0,0)$ with radius $r_0$ (Please be aware that the ball containing $(x_0,0)$ can be centered at any other point but as we are literally working with the usual Euclidean metric, we are allowed to set $(x_0,0)$ as the center). Now, recall $y_{r_0}$ for $r_0$ as mentioned earlier; we have: $$(y_{r_0},0 )\in B((x_0,0), r_0) \cap A, \ \ \ \ \ \ (y_{r_0},0 )\notin q^{-1} (U),$$

which is a contradiction. Hence, $U$ is open in $\mathbb R$ if $q^{-1}(U)$ is open in $A$.

For not being an open map, consider: $$q(\{ (x,y) : x\geq 0 ,y>1 \})= \mathbb R^+ \cup \{0\}.$$

And, for not being a closed map, consider: $$q(\{ (x,\frac{1}{x}) : x>0\})= \mathbb R^+.$$

We are done.

Reza Rajaei
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