Homotopy of Maps

Let I denote the unit interval [0, 1].

Definition 5.1   Let f, g : X$\rTo$Y be two maps. We say that f and g are homotopic if there exists a continuous function

H : X x I$$\rTo$$Y

such that H(x, 0) = f (x) and H(x, 1) = g(x). If A $\subseteq$ X is a subspace such that f and g agree on A, then H is called a homotopy of f and g relative to A or relA if H(a, t) = f (a) for all a $\in$ A and t $\in$ I.

Example 5.2   Let C be a convex subset of Rⁿ and let f, g : X$\rTo$C be any two maps. By definition of convexity the straight line joining any two points in C is contained entirely in C. Thus define H : X x I$\rTo$C by

H(x, t) = (1 - t) ^. f (x) + t ^. g(x).

Then H is a homotopy of f to g. Also if f and g happen to agree on a subspace A of X then this homotopy is relative to A.

Example 5.3   The n-disk Dⁿ in Rⁿ is a convex subset. Thus any two maps f, g : X$\rTo$Dⁿ are homotopic. However, the same statement is far from being true (as we shall see later) if Dⁿ is replaced by its boundary S^{n - 1}. Suppose though that f, g : X$\rTo$S^{n - 1} are two maps such that f (x) $\neq$ - g(x) for all x $\in$ X, i.e. f and g never send a point of X into a pair of antipodal points on the sphere, or equivalently, the straight line segment between f (x) and g(x) never passes through the origin. Then the homotopy defined in the previous example can be divided by its norm to give a homotopy of f to g, i.e.

H(x, t) = $${\frac{(1-t)\cdot f(x) + t\cdot g(x)}{\vert\vert(1-t)\cdot f(x) + t\cdot g(x)\vert\vert}}$$

Example 5.4   Let Y = C $\setminus$ {0} be the punctured complex plane with . Let f : S¹$\rTo$Y be the inclusion of the unit circle. Let g : S¹$\rTo$Y be the unit circle shifted away far enough not to enclose the puncture at the origin, say g(e^2$\pi$it) = 2 + e^2$\pi$it. It is easy to see, at least intuitively that there is no homotopy of f to g. Indeed, in order to deform f into g it will be necessary to ``tear'' the circle, thus destroying continuity.

If f, g : X$\rTo$Y are two homotopic maps then we write f $\simeq$ g.

Theorem 5.5   Let X and Y be spaces. Then homotopy of maps defines an equivalence relation on the set of all maps from X to Y.

Definition 5.6   For spaces X and Y denote by [X, Y] the set of homotopy classes of maps from X to Y.

Example 5.7   Let C be a convex subset of Rⁿ and let X be any space. Then [X, C] is a single point set since any two maps are homotopic.

Homotopy between two maps from X to Y, or ``absolute homotopy'' is a particular case of homotopy relative to a subspace A of X.

Definition 5.8   A topological pair is an ordered pair of spaces (X, A) where A is a subspace of X. If (X, A) and (Y, B) are pairs then a map of pairs f : (X, A) $\rightarrow$ (Y, B) is a continuous function f : X $\rightarrow$ Y, such that f (A) $\subseteq$ B.

Let f, g : (X, A)$\rTo$(Y, B) be maps of topological pairs. We say that f is homotopic to g as maps of pairs, or that f is homotopic to g relative to A if there exists a homotopy

H : X x I$$\rTo$$Y

such that

1.

H(x, 0) = f (x) and H(x, 1) = g(x) namely H is a homotopy from f to g in the usual sense and

2.

H(a, t) $\in$ B for every a $\in$ A and t $\in$ I.

If f and g are maps of pairs as above and f is homotopic to g relative to A, we write f $\sim$ g rel A. One can observe easily that homotopy relative to A is also an equivalence relation. The proof is along the lines of the absolute case.

Exercise 5.9   Show that the relation of relative homotopy is an equivalence relation on the set of maps from (X, A) to (Y, B).

Understanding homotopy classes of maps between two given spaces is in some sense the most fundamental problem in homotopy theory. As one expects from a fundamental problem it is of course a very hard one in general. The concept of induced maps presented above is sometimes of great assistance as it enables us to study a set of homotopy classes of maps we don't understand in terms of one we do.