# Gracious Living

Continuous Maps and Homeomorphisms
November 4, 2010, 16:53
Filed under: Math, Topology | Tags: , , , ,

If you’ve done any calculus, you’ve seen continuous functions.  If you haven’t, the concept isn’t that difficult to understand.  A continuous function is one that doesn’t jump around, instead moving smoothly from point to point.  Formally, we say a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous at a point $x$ if the limits of $f(x_0)$ as $x_0$ approaches $x$ from either side exist and are both equal to $f(x)$.  If the term “limit” is unfamiliar, we can unpack the definition further: $f$ is continuous at $x$ if for every $\epsilon>0$, there is a $\delta>0$ such that whenever $|x-x_0|<\delta$, we have $|f(x_0)-f(x)|<\epsilon$.  This looks pretty nasty, but really all it’s saying is whenever $x_0$ is close to $x$, $f(x_0)$ is close to $f(x)$, and by moving $x_0$ closer, we can get the margin of error as small as we want.

Okay, and now we’ve hit that magic word “close” again, and you know what that means — we can extend this definition to functions between arbitrary topological spaces!

A continuous map between topological spaces is a function $f:X\rightarrow Y$ such that for every open set $U\subset Y$, $f^{-1}(U)$ is open in $X$.  (We don’t expect $f$ to have an inverse: $f^{-1}$ here is the function on $\mathcal{P}(X)\rightarrow\mathcal{P}(Y)$ sending subsets of $Y$ to their preimages in $X$.)

At first blush, the preimage function might seem out of place here.  The statement “for $U\subset X$ open, $f(U)$ is open in $Y$” looks a lot friendlier, and in fact, a function satisfying this condition is called an open map.  But the continuity condition is in fact better for preserving structure.  Its first defense is the epsilon-delta definition given up top for functions on $\mathbb{R}\rightarrow\mathbb{R}$.  Recalling that sets of the form $\{x:|x-x_0|<\delta\}$ form a basis for the Euclidean topology on $\mathbb{R}$, prove that a function on the reals is “topologically” continuous iff it is “analytically” continuous.

Its second defense is the open/closed duality.  If the preimage of every open set is open, then the preimage of a closed set $C$ is $X-f^{-1}(Y-C)$, which is closed.  Meanwhile, we can define a closed map as one that sends closed sets to closed sets, but closed maps are not open maps.  Even the map $f(x)=x^2$ on the reals has $f((-a,a))=[0,a^2)$, which is continuous and closed but not open, while the projection $f(x,y)=x$ on $\mathbb{R}^2\rightarrow \mathbb{R}$ is continuous and open but not closed (the image of a single period of the graph of $\tan x$ is an open interval).

The third defense is the neighborhood viewpoint: a function between topological spaces is continuous at $x$ if and only if every neighborhood of $f(x)$ includes some $f(U)$ for $U$ a neighborhood of $x$.  In particular, if $x$ is a limit point of $A$, then $f(x)$ is a limit point of $f(A)$.  So if a sequence converges to $x$ in the domain, then its image converges to $f(x)$ in the range, and in general, $f(\overline{A})\subset \overline{f(A)}$.  The same cannot be said about interiors: what can?

We talked about ordering the possible topologies of a space a while back.  Prove that if we have a continuous function $f:X\rightarrow Y$, then it stays continuous if we give $X$ a finer topology or $Y$ a coarser one.  If $X$ is discrete or $Y$ is indiscrete, then any map between the two is continuous!  (If $Y$ is discrete or $X$ is indiscrete, then any map is open and closed.)

Prove that the composition of two continuous functions is continuous.  This is so insanely useful, it’s not even funny.

Several of the topologies we’ll see in the future are constructed by defining certain functions to be continuous.  Imagine that we have a function $f:X\rightarrow Y$, where $X$ has a topology but $Y$ is just a set.  Then we can place a topology on $Y$ by saying “a set is open iff its preimage is open in $X$“; this is the finest topology that makes $f$ continuous (if we added another open set to $Y$, its preimage would not be open in $X$).  Similarly, if $Y$ has a topology and $X$ doesn’t, we can place a topology on $X$ by saying “a set is open iff it is the preimage of an open set in $Y$“; this is the coarsest topology on $X$ that makes $f$ continuous.

I want to say something about continuity of functions on the reals, since this really confused me when I was learning topology.

In school, they often classify the different types of discontinuities for real-valued functions into something like the above.  “Removable” discontinuities are when the limit exists, but the function has no value; “jump” discontinuities are when the limits are different from each side; and “essential” discontinuities are when one of the limits is infinite or does not exist.1 It’s easy to see that the second and fourth functions are topologically discontinuous, but I wracked my brains over functions of the first or third type.  The solution to the problem is that the first and third functions aren’t actually functions!  At least, not on $\mathbb{R}$, since they aren’t defined at a point.  If we define them to have some value at that point, they become discontinuous (as long as the value doesn’t fill the “hole” in the first one).  Alternatively, we could consider the domain to be $\mathbb{R}$ minus a point, in which case they are continuous.

Continuous functions are, in fact, the morphisms of the category of topological spaces.  Morphisms are like half an isomorphism: they preserve structure but aren’t reversible, so they don’t give you an equivalence relation.  In this case, the full isomorphism is called homeomorphism, and it is pretty easy to define: a homeomorphism is a continuous bijection whose inverse is also continuous.  (Or just a continuous open bijection.)  So a homeomorphism not only matches up the sets, it also matches up their topologies: every open set in one space will have an open image or preimage in the other one.  Since the inverse of a homeomorphism is also a homeomorphism, the “preimage/image” distinction doesn’t really matter, either.  In general, we might not define an explicit homeomorphism, instead just saying that two spaces are homeomorphic.

In a way, the topology we’re doing is the study of spaces up to homeomorphism.  That is, we’re only concerned about the things that are preserved by homeomorphism (the topology and its consequences), as opposed to the extrinsic characteristics of a space that might be more visible.

Here’s an example.  In the post on countability, we used a stereographic projection to shrink the real line down to a semicircle, then we projected down to the open interval $(0,1)$.  (Real stereographic projection is done with lines through the north pole of the circle, so the image of the real line is the circle minus its north pole; you can then “unwrap” this set to give you an open interval in the line.)  You should check that this function (either version) is continuous, either by defining an appropriate topology on the semicircle and checking both stages, or just looking at the preimage of an open set in $(0,1)$.  But notice also that it has a natural inverse, given by running the steps backwards: first send the interval up to the circle or semicircle, then project it out to the real line.  This is also continuous.

So the real line and open interval $(0,1)$ are homeomorphic — in fact, since there’s a linear map between any two open intervals, any open interval is homeomorphic to $\mathbb{R}$.  (And this isn’t the only way — the tangent function is a homeomorphism between $(-\pi/2,\pi/2)$ and $\mathbb{R}$.)  What this means is that the property of boundedness is an extrinsic property: though the interval looks “small” and the line looks “infinite,” they are topologically identical.  We will soon be talking about the compactness property, which is a topological way of thinking about boundedness.  Similarly, since both are homeomorphic to the circle minus a point, the property of curvature is also extrinsic.  I can’t think of any topological equivalent to this one, though it is very important in the more rigid study of differential geometry.

I’ll try to post again later today, since this should’ve been up yesterday.  Probably back to set theory.

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Let’s not harbor any illusions here. Discontinuity can look a lot worse than this. Math is dangerous.