# Mathematics Prelims

## March 12, 2009

### Homotopy of Functions

Filed under: Topology — cjohnson @ 6:51 pm

Given two continuous real-valued functions, it’s easy to imagine how one of these functions may be deformed into the other.  For instance, if we take the sine curve modified so that it passes through the points (-1, 0) and (0, 1), we can bend and stretch it into a curve that passes through (-1, 0) and (0, 1) but as a semicircle.  Consider the video below.

Here we’re taking the curve, grabbing the endpoints and swapping them, bending the curve as we do so.  In topology this idea of deforming one function into another is formalized by homotopy.  A homotopy between two functions, $f$ and $g$, is a continuous map that “through time” bends one curve into the other.  If a homotopy between $f$ and $g$ exists we say that $f$ and $g$ are homotopic and write $f \simeq g$.

(It should be noted that the above deformation is with parametric curves.  That is, these are functions of the form ${}[0, 1] \to \mathbb{R}^2$, and not $\mathbb{R} \to \mathbb{R}$ functions.  This is why having the crossing and failing the vertical line test is okay.)

Formally, a homotopy between two continuous functions $f, g : X \to Y$ is a continuous function $F : X \times I \to Y$ that satisfies a few properties.  Before discussing those properties, however, let’s make some ideas and notation clear.  First we’ll just let $I$ denote the closed unit interval [0, 1], mainly just to keep typing down.  Now, when we say a continuous function from $X \times I$ to $Y$, we of course mean continuous in the topological sense (preimages of open sets are open), but what’s an open set in $X \times I$?  It’d be nice to say that it was just the Cartestian product of an open set in $X$ and an open set in $I$, but unfortunately that’s not enough.

If you think about Euclidean plane, $\mathbb{R}^2$, it seems entirely reasonable to say that the ball $B_1(0) = \{ (x, y) \in \mathbb{R}^2 : \|(x, y)\| < 1\}$ (that is, the collection of all points of distance less than one from the origin) is open, but we can’t represent this ball as the product of two open sets in the real line.  If we take an open set $A$ and an open set $B$ from $\mathbb{R}$, their product $A \times B$ is just going to be a collection of open rectangles in the plane.  Taking $A = B = (1, 2) \cup (3, 4)$, for instance, we have the following.

If $A$ and $B$ are open sets, then $A \times B$ is going to look something like this, since every open set can be represented as a countable collection of open intervals.  Of course, we could construct a more complicated set by changing the sizes and number of intervals making up $A$ or $B$, but we’re still limited to sets that look something like the above.

The open ball, however, can not be represented like this.  If we took lots and lots, infinitely many in fact, open rectangles of various sizes and could put them together however we like, we could construct the open ball.  That is, the open rectangles of $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$ should form a basis for the topology.

The product topology of $X \times Y$ is precisely the topology whose basis is the collection of the open “rectangles” in $X \times Y$.  So, even though an open set in $X \times Y$ may not be the product of an open set in $X$ and an open set in $Y$, it can be described as a union of those open rectangles (where we allow infinite unions).  It is in this sense that we mean $F : X \times I \to Y$ is continuous: the preimage of an open set in $Y$ is an open set in $X \times I$ with the product topology.

We may think of the function $F : X \times I \to Y$ as actually being a family of functions $\{ f_t : X \to Y \}$, for $t \in [0, 1]$ where $f_t$ actually means the function $f_t(s) = F(s, t)$.  We will generally think of homotopies as deforming one function into another through time, where “time” is given by the interval [0, 1].  That is, the starting time is $t = 0$ and the stopping time is $t = 1$.  At each point $t \in [0, 1]$  associate the function $f_t$.  So, the homotopy is giving us a family of functions, one for each moment in time.  The condition of continuity basically means that if $\epsilon > 0$ is small, then the change from $f_t$ to $f_{t + \epsilon}$ is small.

Now, earlier we said that this function $F : X \times I \to Y$ has to satisfy a few properties to be a homotopy taking $f$ to $g$.  The first condition was continuity, and the second is simply that $F(s, 0) \equiv f_0 \equiv f$ and $F(s, 1) \equiv f_1 \equiv g$.  This means that when we start, we have the function $f$ and when we finish we have the function $g$.  So, a homotopy gives us a way to take a function $f$ and bend it around however we like provided that we don’t break the curve and that we eventually stop bending the function when we get to the function $g$.

In the special case of the $\mathbb{R}^n$ valued functions we can actually show that all continuous maps are homotopic to one another.  What we will do is add little pieces of $f$ to little pieces of $g$, where the portion of $f$ that we’re using shrinks, and the portion of $g$ grows, as $t$ goes from 0 to 1.  We then define the homotopy  $F(s, t) = (1 - t) \cdot f(s) + t \cdot g(s)$.  It’s easy to see that this satisfies the conditions for a homotopy as when $t = 0$ we have $F(s, 0) = f(s)$, and when $t = 1$ we have $F(s, 1) = g(s)$.  Since this is composition of continuous functions (addition, substraction, multiplication, and the maps $f$ and $g$), this is also a continuous function.  This is known as the straight-line homotopy since we imagine taking the curve $f$ and the curve $g$ and drawing lines between associated points (i.e., drawing a line between f(s) and g(s) for each s), then the homotopy just moves $f$ to $g$ along those lines.  This result applies to convex subspaces of $\mathbb{R}^n$ as well.

At this point you may wonder about how to find two functions that are not homotopic.  We’ve just seen that all continuous $\mathbb{R}^n$ functions are homotopic to one another, so we need to consider some other space.  Suppose we just take $\mathbb{R}^2$ and remove the origin; $\mathbb{R}^2 \setminus 0$.  If $f : \mathbb{R} \to \mathbb{R}^2 \setminus 0$ is the unit circle, and $g : \mathbb{R} \to \mathbb{R}^2 \setminus 0$ is a line segment connecting the points (2, -1) and (2, 1), then we can’t deform the circle into this line segment in our space.

Intuitively it’s easy to see that in order to deform the circle like this we’d either have to tear the circle, which would make the deformation discontinuous, or the circle would have to pass through the origin as it’s deformed.  However, the origin isn’t in our space, so there can be no intermediate function in our homotopy that passes through the origin.  This means the circle and the line segment can’t be homotopic.

1. […] essentially the same thing or not.  That is, if one path can be deformed into the other.  We saw last time that homotopy gave us a way to deform functions, but we have to take a little bit of care when it […]

Pingback by Path Homotopy « Mathematics Prelims — March 15, 2009 @ 12:11 pm

2. Hi: you wrote:

“Since this is composition of continuous functions (addition, substraction, multiplication, and the maps and ), this is also a continuous function”

however, I worry that this is not true: given f: X -> Z, g: Y -> Z, it does not always follow that the pair f,g:X x Y -> Z is continuous (see Munkres chapter 18). X is our original space, and Y is the [0,1] time domain. Please clear this up for me.
Thanks

Comment by Alex Moll — October 23, 2009 @ 8:22 pm

3. […] Homotopy of Functions […]

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