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I was going through an old topology prelim, and encountered a question which I'm really not sure how I should work out. Here it is:

Suppose we let $X = \mathbb{R} \times \{3,4,…\} \subset \mathbb{R}^2$. Now let $L_{\theta} \subset \mathbb{R}^2$ be the line through the origin with slope $\tan \theta$, i.e. the directed angle from the positive $x$-axis to $L_{\theta}$ is $\theta$. Further, we let $$ Y= \bigcup_{i \geq 3} L_{\pi/i}.$$ Also, we define $g: X \rightarrow Y$ by $g(x,i)= (x, x\tan(\pi/i))$. We have to show that $g$ is a continuous surjection, but not a quotient map. Any ideas how I should approach this?

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1 Answer 1

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Proving that $g$ is a continuous surjection is straightforward. Note also that $g$ is almost a bijection: the origin is the only point of $Y$ that has more than one pre-image under $g$. Thus, you can expect that this point will be significant in showing that $g$ is not a quotient map.

To show this, let $$A=\left\{\left\langle x,x\tan\frac{\pi}k\right\rangle\in Y:|x|<\frac1k\right\}\;,$$

and show that $A$ is not open in $Y$, but $g^{-1}[A]$ is open in $X$.

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Just for my understanding, is there a theorem that characterizes quotient maps in the Euclidean setting by being open? Because for general topological spaces being open is not necessary for a continuous surjection to be a quotient map. –  erlking Dec 15 '12 at 0:50
@erlking: No, that’s not the point at all. The point is that by definition a set is open in the quotient topology iff its preimage is open in the original space. –  Brian M. Scott Dec 15 '12 at 0:52
I don't understand the inequality $|k| < \frac{1}{k}$. I believe $k$ is an integer greater than or equal to $3$, and that is why I am somewhat confused. Can you please clarify that? –  Libertron Aug 14 '13 at 23:32
@Libertron: It’s a fairly obvious typo: it should be $|x|<\frac1k$. I’ve fixed it now. –  Brian M. Scott Aug 14 '13 at 23:38
@Libertron: No, it does not mean showing that $g$ is continuous. It means only what it says: show that the set $g^{-1}[A]$ is open in $X$. Do this by seeing exactly what the set $g^{-1}[A]$ is, and verifying that it is indeed open in the space $X$. –  Brian M. Scott Aug 15 '13 at 16:48

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