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I am looking for easy(!) counterexamples that the product of two quotient maps is not necessarily a quotient map and that the quotient space of a Hausdorff space is not necessarily Hausdorff.

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For the second example start with $\Bbb R$ with the usual topology, and let $X$ be the quotient space obtained by identifying the set of negative reals to a point $p$, so that $X=\{p\}\cup[0,\to)$. Then $X$ isn’t even $T_1$, since every open nbhd of $0$ in $X$ contains $p$.

I don’t know a really simple example of the first type. Engelking has the following example. Let $$X_0=Y_0=\Bbb R\setminus\left\{\frac1n:n\in\Bbb Z^+\right\}$$ with the topology inherited from $\Bbb R$. Let $X_1=\Bbb R$, and let $Y_1$ be the quotient of $X_1$ obtained by identifying the set of positive integers to a point. Let $f_0:X_0\to Y_0$ be the identity map, and let $f_1:X_1\to Y_1$ be the quotient map. Let $f:X_0\times Y_0\to X_1\times Y_1$ be the product map.

For $k,\ell\ge 2$ let

$$F_{k,\ell}=\left\{x\in X_0:\left|\frac1\ell-x\right|\le\frac1k\right\}\times\left\{\ell-\frac1k\right\}\subseteq X_0\times X_1\;;$$

clearly each $F_{k,\ell}$ is closed in $X_0\times X_1$. It’s not hard to check that $\mathscr{F}=\{F_{k,\ell}:k,\ell\ge 2\}$ is a locally finite collection and therefore closure-preserving, so $F=\bigcup\mathscr{F}$ is closed in $X_0\times X_1$. Finally, note that $\langle 0,f_2(1)\rangle\in(\operatorname{cl}f[F])\setminus f[F]$, so $f[F]$ is not closed in $Y_0\times Y_1$, while $f^{-1}[f[F]]=F$ is closed in $X_0\times X_1$; this shows that $f$ cannot be a quotient map.

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    $\begingroup$ The prodigal son returns. Welcome back, Brian. $\endgroup$ – Asaf Karagila Oct 4 '14 at 21:06
  • $\begingroup$ @Asaf: Thanks. My activity will likely be pretty limited, at least for a while, but I am at least keeping an eye on general topology. $\endgroup$ – Brian M. Scott Oct 4 '14 at 21:32
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Might be an easier example for the first one from Munkres: Let $K = \{\frac {1} {n} | n \in \mathbb Z_+\}$ and $\mathbb R_K$ be the space with underlying set $\mathbb R$ and topology generated by basis $\{(a,b)| a,b \in \mathbb R\} \cup \{(a,b)-K| a,b \in \mathbb R\}$. Munkres calls it the K topology.

Now let Y be the space obtained from $\mathbb R_K$ by collapsing $K$ to one point. Let $p: \mathbb R_K \to Y$ be the quotient map. Note that Y is clearly $T_1$ but not Hausdorff since any neigbourhood of $0$ in $\mathbb R_K$ intersects with K. So the diagonal $\Delta _Y $ is not closed in $Y\times Y$. But $(p\times p ) ^{-1}(\Delta _Y) = \Delta_{\mathbb{R}_K} \cup K\times K$ is closed. So, $p\times p$ is not quotient.

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  • $\begingroup$ @Eric It isn't clear to me why p is a quotient map. Can you please explain in short? $\endgroup$ – User Not Found May 14 '16 at 4:50
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    $\begingroup$ @ArghyaChakraborty: $p$ is defined to be a quotient map. The topology on $Y$ is defined to be the quotient topology. $\endgroup$ – Eric Wofsey May 14 '16 at 4:52
  • $\begingroup$ @EricWofseyI guess my knowledge isn't clear on how quotient maps are defined automatically. Can you suggest a pdf or web page to me for getting a clear concept. Thanks. $\endgroup$ – User Not Found May 14 '16 at 4:56
  • $\begingroup$ @EricWofsey Ok I got it thanks a lot. $\endgroup$ – User Not Found May 14 '16 at 5:00

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