Let $X = \Bbb{R} \times \{3, 4, ...\} \subset \Bbb{R}^2$. Let $L_{\theta} \subset \Bbb{R}^2$ be the line through the origin with slope $\tan \theta$ (i.e., the directed angle from the positive $x$-axis to $L_{\theta}$ is $\theta$), and let $$Y = \bigcup_{i \geq 3} L_{\pi/i}.$$ Define $g: X \rightarrow Y$ by $g(x,i) = (x, x \tan(\frac{\pi}{i}))$.Show that $g$ is a continuous surjection but not a quotient map.

By the universal mapping property, $g$ is continuous if an only if $g_1$ and $g_2$ are continuous, where $g_1 : X \rightarrow \Bbb{R}$ with $g(x,i)=x$ and $g_2: X \rightarrow Y'$ (where $Y'$ is the set of $y$-coordinates in $Y$) with $g(x,i) = x \tan(\frac{\pi}{i})$. We know that $g_1$ is obviously continuous because it is a projection. $g_2$ is continuous because both the tangent function and $\frac{\pi}{i}$ are continuous, and the composition of continuous functions is continuous. Finally, multiplying it be $x$ does not affect its continuity because the product of two continuous maps is continuous, implying that $g$ is continuous.

For any point $(x, y) \in Y$, if $x=0$, then the preimage is $(0,i)$ for any $i \in \{3, 4, ...\}$. If $x \not= 0$, then the preimage is $(x, (\arctan(\frac{y}{x})\pi )^{-1})$. So $g$ is surjective.

We can make $g$ into an injective function if we identify the points $(0,i)$ and $(0,j)$ for all $i \in \{3, 4, ... \}$. So we have the following commutative diagram,

(for some reason I could not draw the diagram, but it is a diagram with $g: X \rightarrow Y$, $\bar{g}: X/ \sim \rightarrow Y$ where $\sim$ is the equivlence relation defined above, and $p: X \rightarrow X/ \sim$)

where $p$ is obviously a quotient map. By the universal mapping property for quotients, $g$ is a quotient map if and only if $\bar{g}$ is. Suppose, for contradiction, that $g$ is a quoteint map.

Since the only preimage of $(0,0) \in Y$ under $\bar{g}$ is $(0,i)$ for $i \in \{3, 4, ...\}$, $\bar{g}$ is injective. But then $\bar{g}$ must be a homeomorphism since it is an injective quotient map. This means that $\bar{g}$ is open. However, the set $U = (\Bbb{R} \times \{3\}) \cap (\Bbb{R} \times \{4\})$ is open in $X/ \sim$ while $g(U) = \{0\}$ is closed in $Y$. A contradiction. Therefore, both $\bar{g}$ and $g$ cannot be quotient maps.

Is my proof correct?

  • $\begingroup$ Your stuff is missing the "following commutative diagram"... $\endgroup$
    – DonAntonio
    May 30 '14 at 15:25
  • $\begingroup$ @DonAntonio Yes, there was a problem with my computer...I tried to fix it but I couldn't upload the image so I just described the diagram. $\endgroup$
    – Artus
    May 30 '14 at 15:26
  • $\begingroup$ That's just enough for anyone understanding the basic theory in this, @Artos. +1 $\endgroup$
    – DonAntonio
    May 30 '14 at 15:27
  • $\begingroup$ It looks fine, @Artos, yet one question: did you even try to prove $\;g\;$ is not an open map? That'd prove it is not a quotient map and, perhaps, that'd be somehow easier... $\endgroup$
    – DonAntonio
    May 30 '14 at 15:29
  • $\begingroup$ @DonAntonio But not all quotient maps are open, right? So I tried to looked at $\bar{g}$ instead because that is supposed to be a homeomorphism, which must be open (but is not). $\endgroup$
    – Artus
    May 30 '14 at 15:31

Let $V = \cup_{n \ge 3} \{ (x,x \tan { \pi \over n}) \,| \, |x| < {1 \over n} \} \subset Y$, and $U \ g^{-1}(V) = \cup_n (-{1 \over n}, {1 \over n}) \times \{n \} \subset X$.

$U$ is open, but $V$ is not.

The following was my previous, incorrect answer.

Note that any open set in $Y$ that contains $(0,0)$ must contain some segment of all the $L_{\pi \over i}$.

Let $U = (-1,1) \times \{3\}$ which is open in $X$. However, $g(U) \subset L_{\pi \over 3}$, hence $g(U)$ cannot be open.

  • $\begingroup$ For some reason I got confused about this again. We need to find $V \subseteq Y$ not open with $g^{-1}(V) \subseteq X$ open, right? However, if we let $V = g(U)$, then $g^{-1}(g(U)) = \{(x,i) \in X : g(x,i) \in L_{\pi/3} \} = [(-1,1) \times \{3\}] \cup [\{0\} \times \{4\}] \cup [\{0\} \times \{5\}] \cup ... $, which is not open. $\endgroup$
    – Artus
    Jun 10 '14 at 17:46
  • $\begingroup$ I have fixed my answer, thanks for catching this. $\endgroup$
    – copper.hat
    Jun 10 '14 at 19:11

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