Let $(Y, \tau)$ be a topological space and let $X$ be a set such that there exist a surjective function $f \colon X \to Y$.

Consider $\tau_1$ the smallest topology in $X$ that makes $f$ a quotient map,

$\tau_2$ the smallest topology $X$ that makes $f$ continuous,

$\tau_3$ the smallest topology in $X$ that makes $f$ an open map,

$\tau_4$ the smallest topology in $X$ that makes $f$ a closed map,

$\tau_5$ the smallest topology in $X$ that makes $f$ an open and closed map,

$\tau_6$ the smallest topology in $X$ that makes $f$ a closed and continuous map.

I am asked to compare each topology, I understand the definition for "the smallest topology" but it still troubles me, because I don't know how to use it. Can someone give me an example of comparing these topologies?

  • $\begingroup$ Your $\tau_1$ and $\tau_2$ seem to be the same thing; was one of them supposed to be something else? $\endgroup$ – Brian M. Scott Dec 12 '11 at 22:10
  • $\begingroup$ i corrected. $\tau_1$ is for quotient map $\endgroup$ – Alex J. Dec 12 '11 at 22:12

If $\tau_1$ and $\tau_2$ are topologies on the same set $X$, we say that $\tau_1$ is smaller or coarser than $\tau_2$ if $\tau_1\subseteq\tau_2$. (Note that this terminology is a little sloppy, since $\tau_1$ is allowed to be equal to $\tau_2$. If you want to specify that $\tau_1\subsetneqq\tau_2$, it’s best to say that $\tau_1$ is strictly weaker or coarser than $\tau_2$.

For a simple example, let $\tau_1$ be the usual Euclidean topology on $\mathbb{R}$, and let $\tau_2$ be the discrete topology on $\mathbb{R}$; since $\tau_2=\wp(\mathbb{R})$, it’s clear that $\tau_1\subseteq \tau_2$, so the Euclidean topology on $\mathbb{R}$ is weaker (or coarser) than the discrete topology. (In fact it’s obviously strictly weaker.) On the other hand, if we let $\tau_3=\{\mathbb{R}\setminus F:F\text{ is finite}\}$, the cofinite topology on $\mathbb{R}$, then $\tau_3\subsetneqq\tau_1$: the cofinite topology is strictly weaker than the Euclidean topology.

Added: A useful fact is that if $\mathscr{S}$ is any collection of subsets of $X$, and $\tau$ is the intersection of all of the topologies on $X$ that contain $\mathscr{S}$, then not only is $\mathscr{S}\subseteq\tau$, but $\tau$ is again a topology on $X$. (This is an easy exercise using the definition of a topology.) Clearly, then, $\tau$ is the weakest (smallest, coarsest) topology on $X$ that contains every member of $\mathscr{S}$, which therefore always exists.

This means that if you have some collection $\mathscr{S}$ of sets that you want to be open, you can find the weakest (smallest, coarsest) topology that does the trick by intersecting all of the topologies that make the members of $\mathscr{S}$ open. The collection $\mathscr{S}$ is said to be a subbase for the resulting topology.

To return to the specifics of your question, notice that the more restrictions you put on $f$, the more open sets you’ll require in $X$. For instance, $\tau_5$ makes $f$ both open and closed, so it automatically makes $f$ open. Thus, it’s one of the topologies that are intersected to find $\tau_3$, and as a result $\tau_3\subseteq\tau_5$.

  • $\begingroup$ That's what I thought about $\tau_5$ and $\tau_3$. Does it follow, say, since a quotient map is continuous, so $\tau_2 \subset \tau_1$? $\endgroup$ – Alex J. Dec 12 '11 at 22:24
  • $\begingroup$ @Alex: Yes, it does: $\tau_1$ is one of the topologies that are intersected to find $\tau_2$. $\endgroup$ – Brian M. Scott Dec 12 '11 at 22:32

$\tau_3 = \tau_4 = \tau_5 = \{ \emptyset, X \}$, the indiscrete topology on $X$. This is clear, as the indiscrete topology is the smallest topology on $X$ anyway and any surjection is both an open and closed map under this topology ($f[\emptyset] = \emptyset$, and $f[X] = Y$, and both sets are open in $Y$).

Assuming $\tau_6$ is well-defined, this is a topology that makes $f$ a closed and continuous function, and so a quotient map. So $\tau_1$, as the smallest such topology obeys $\tau_1 \subset \tau_6$, and as $\tau_1$ makes $f$ continuous in particular, and $\tau_2$ is the smallest such topology, again we conclude $\tau_2 \subset \tau_1 \subset \tau_6$. Equality is possible, e.g. when we have $f$ a bijection. The equal (see above) $\tau_3, \tau_4, \tau_5$ are a subset of all of them, of course.

The smallest topology satisfying some property $\mathcal{P}$ is simply, as others noted too, the intersection of all topologies on the set that satisfy $\mathcal{P}$, and as the intersection of topologies is always a topology, the well-definedness comes down to finding or defining just one topology on the set $X$ that obeys $\mathcal{P}$. But I'm not yet sure whether there always exists a topology on $X$ that makes $f$ closed and continuous for a given $f$. Continuity is easy: we start with all inverse images of open sets of $Y$. This gives us new closed sets, the image of which must be closed, but the topology on $Y$ is given and cannot be controlled... So there probably are examples of functions such that $f[X\setminus f^{-1}[O]]$ is not open for some open subset $O$ of $Y$, and then we cannot have any topology that makes $f$ closed and continuous and this makes $\tau_6$ ill-defined in a general setting, I think. But like I showed, if it exists, it's the biggest of the considered topologies. [edit] come to think of it, is there always a topology on $X$ to make $f$ quotient? I'm not aware of a construction, so I have my doubts there as well.

  • $\begingroup$ You don’t need $f[X\setminus f^{-1}[O]]$ to be open: you need it to be closed, which it is, since it’s $Y\setminus O$. (Remember, $f$ is surjective.) Giving $X$ the topology $\{f^{-1}[U]:U\text{ is open in }Y\}$ automatically makes $f$ continuous, open, closed, and a quotient map. $\endgroup$ – Brian M. Scott Dec 16 '11 at 18:36

the set of all topologies on $X$ can be partially ordered by inclusion (as subsets of $2^X$), $\tau\leq\tau'$ iff $U\in\tau\Rightarrow U\in\tau'$. for instance, the smallest topology on $X$ is $\{\emptyset,X\}$ and the largest is $\{S | S\subseteq X\}$. if $f$ is to be continuous, then $\tau_1$ must contain all sets of the form $f^{-1}(U), U\in\tau$. as for "smallest," if $\mathcal{S}\subseteq 2^X$ there is a unique smallest topology containing $\mathcal{S}$, namely the intersection of all topologies containing $\mathcal{S}$ (exercise). it seems to me that the smallest such that $f$ is open/closed/clopen is the indiscrete topology (this uses surjectivity of $f$)


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