2
$\begingroup$

I was studying some basic topology and was wondering about the basis for a quotient topology. I came up with the following, but I can't seem to find something this elementary stated explicitly anywhere, and this set of notes (link is 404, only available in Google cache) seems to suggest that a the (fix: Brian) basis is in general unclear. Have I made a mistake somewhere?

Lemma: Let $q: X\rightarrow X/{\sim}$ be a quotient map, and let $\mathcal{B}$ be a basis for the original topology $\tau$ on $X$. Then $q(\mathcal{B})=\{q(B)\mid B\in\mathcal{B}\}$ is a basis for the quotient topology $\tau_q$ iff $q$ is an open map.

Proof: Suppose $q(\mathcal{B})$ is a basis for $(X/{\sim}, \tau_q)$. Therefore, the image of each $B\in\mathcal{B}$ is open, and since every open $O\subseteq X$ is a union of the basis elements (and $q(\emptyset)=\emptyset$), $q(O)$ is open. Thus $q$ is an open map.

Suppose $q$ is open. $q$ is surjective, so $q(\mathcal{B})$ covers $X/{\sim}$ since $\mathcal{B}$ covers $X$. Now, consider the intersection $O':=q(B)\cap q(B')$. It is open, since $q$ is open. Thus $O:=q^{-1}(q(B)\cap q(B'))$ is open as $q$ is continuous. Since $q$ is surjective, $O'=q(q^{-1}(O'))=q(O)=\bigcup_{\delta}q(B_{\delta})$ as $O$ is a union of some basis elements $B_{\delta}$. So for all $x\in O', x\in q(B_{\delta})\subseteq O'$ for some $\delta$. Thus $q(\mathcal{B})$ is a basis. By the same construction, any open set in $\tau_q$ is expressible as a union of basis elements, and conversely any union of basis elements is in $\tau_q$. Thus $q(\mathcal{B})$ is indeed a basis for $\tau_q$. $\blacksquare$

Edit: I just realized this certainly does not solve for the general case, derp. It still seems quite useful to not appear though. For instance, the typical question of the quotient of $\mathbb{R}^2$ by $(x,y)\sim(a,b)\Leftrightarrow x^2+y^2=a^2+b^2$ can be seen to be an open map, since every 'open box' in $\mathbb{R}^2$ maps to open intervals in $[0,\infty)$, so the intervals $[0, x)$ and $(a, b)$ with $a$ positive form a basis.

  • 0
    There are a couple of minor infelicities in the write-up, but you’ve made no mistake. However, it *is* an error to speak of *the* base for a topology: topologies have more than one base. – 2017-01-03
  • 0
    Did I do that? Thanks for catching it. – 2017-01-03
  • 0
    link is a 404 BTW – 2017-01-05
  • 0
    @HennoBrandsma The page seems to have disappeared, it is currently still available from [Google's cache](http://webcache.googleusercontent.com/search?q=cache:J8jOmzwSgU8J:math.umn.edu/~sglasman/5345hf16/5345h0921.pdf+&cd=4&hl=en&ct=clnk&gl=us&client=firefox-b-ab), not sure how long it will be there though. The sentence I am talking about is on page 3 (search for 'basis') – 2017-01-05
  • 0
    The quotient topology on $X/\sim$ is defined to be the strongest topology on $X/\sim$ such that $q$ is continuous . So it is equal to $\{q(t):t\in \tau\}$ and $q$ is an open map. – 2017-01-05
  • 0
    @user254665 Not all quotient maps are open, for example, the equivalence relation $0 \sim 2$ on the topology $X=\{0, 1, 2\}$ with $\tau=\{\emptyset,\{0\},\{1,2\},X\}$ does not induce an open map, as the open sets in $X/{\sim}=\{[0], [1]\}$ are $\tau_q=\{\emptyset,X/{\sim}\}$. Then $q(\{0\})=\{[0]\}$ which is not open. – 2017-01-05
  • 0
    reading the notes, I don't see how your lemma appears useful there. It just uses the definition, no more. And in this example, the quotient map is not even open, but closed and not open. – 2017-01-05
  • 0
    In your example with $X=\{0,1,2\}$ we have $q(X)=\{ \{0,2\},\{1\}\}.$ The members of $q(X)$ are not even members of $X.$ How do wish to define the topology on $q(X)$? – 2017-01-05
  • 0
    @user254665 The quotient map $q$ sends $x\in X$ to the equivalence class $[x]\in X/{\sim}$, so the members of $q(X)\,(=X/{\sim})$ are necessarily not members of $X$ (since they are equivalence classes of elements in $X$. I believe the quotient topology $\tau_q$ on $X/{\sim}$ is then defined by $\tau_q:=\{U\subseteq X/{\sim}\,|\, q^{-1}(U)\in\tau\}$, i.e. all subsets of $X/{\sim}$ whose preimage is open in $X$. In the example in my comment the non-trivial proper subsets of $X/{\sim}$ are $\{[0]\},\{[1]\}$ and neither of those have open preimages so the quotient topology is the trivial topology. – 2017-01-06
  • 0
    @HennoBrandsma I didn't actually really read the notes, I was just using it as a source that says that in general finding a basis for the quotient topology is hard (which it probably is), and I wasn't trying to apply the lemma there. I'm not exactly sure which example you are referring to though, could you be more specific? – 2017-01-06
  • 0
    See Engelking, General Topology, chapter 2, first paragraph of section 2.4 for a def'n of quotient topology. – 2017-01-06
  • 0
    @user254665 Engelking, definition of quotient topology: "... denote by $X/E$ the set of all equivalence classes of $E$ and by $q$ the mapping of $X$ to $X/E$ assigning the point $x\in X$ the equivalence class $[x]\in X/E$. ... in the class of all topologies on $X/E$ that make $q$ continuous there exists the finest one: this is the family of all sets $U$ such that $q^{-1}(U)$ is open in $X$." Both our definitions (construction from inverse mapping, and characterization by finest topology) are equivalent, but your construction ($\tau_q=q(\tau)$) is wrong as it has too many open sets. – 2017-01-07

0 Answers 0