Homotopies

Definition 1.1
Let

f:X⟶Y
and

f′:X⟶Y
be two continuous maps of the space X into the space Y. We say that f is homotopic to f′, written f≃f′, if there exists a continuous map

F:X×I⟶Y
where I denotes the interval [0,1], such that

F(x,0)=f(x)
and

F(x,1)=f′(x)
The map F is called a homotopy between f and f′.

If f≃f′ and f′ is a constant map, we call f null-homotopic.

In other words, two functions f and f′ are homotopic if one can be continuously transformed into the other, and F represents that transformation. If we think of the second parameter of F as time, at t=0 F is equal to f. As time increases, F begins to morph and by t=1, it is equal to f′

A famous example of a homotopy is between the coffee cup and the donut ortorus,mathematicallyspeaking If we imagine a coffee cup made of clay, we can mold it into a donut shape without creating a new hole whichwouldrepresentadiscontinuity.

Definition 1.2
Let f and f′ be two paths

f:X⟶X

f′:X⟶X
mapping the interval I into X, with endpoints x0 and x1. f and f′ are said to be path homotopic, written f≃Pf′, if there exists a continuous map

F:I×I⟶X
such that, given any s,t∈I

F(s,0)=f(s),F(s,1)=f′(s)
and

F(0,t)=x0,F(1,t)=x1
We call F a path homotopy between f and f′.

We can think of a path homotopy as a more specific version of a homotopy. We know that a path in topology traces a curve between two fixed endpoints. The second condition specifies that the endpoints x0 and x1 should say the same even as t increases.

All path homotopies are homotopies, but not the other way around!

Lemma 1.3
The relations ≃ and ≃P are equivalence relations.

Proof: We need to show that the relations ≃ and ≃P are reflexive, symmetric, and transitive. For the first two, we will look at ≃ only, leaving out ≃P since it will follow by the same argument. For transitivity, we will need to address the extra condition of ≃P which requires that the endpoints stay the same.

    


Reflexivity: Let

f:X⟶Ybe a continuous map.

f≃f with the trivial homotopy

F(x,t)=f(x)∀t∈[0,1]

Symmetry: Define f as before and let Let

f′:X⟶Y
be another continuous function such that f≃f′. Let F(x,t) be a homotopy between f and f′, meaning that

F(x,0)=f(x)andF(x,1)=f′(x)
Then, we can define G(x,t)=F(x,1−t), which is also a homotopy between f and f′. We can think of G as F going backwards, since t is starting at one and going to zero. So we have

G(x,0)=F(x,1)=f′(x)
and

G(x,1)=F(x,0)=f(x)
which gives us
f≃f′.

Transitivity: Introduce a third path

f″:X⟶Ysuch that f≃f′ and f′≃f″. Let F be a homotopy between f and f′ and let F′ be a homotopy between f′ and f″. We need to find a homotopy between f and f″.
Define

G:X×I⟶Y
by

G(x,t)={F(x,2t)F′(x,2t−1)t∈[0,12]t∈[12,1]
While t only goes from zero to one, this transformation takes place over 2 “seconds”, since its components go from 0 to 1 one after another. It’s clear that this maps f to f″, since it does the transformation of F followed by the transformation of F′. We can even check:

G(x,0)=F(x,2⋅0)=F(x,0)=f(x)

G(x,1)=F′(x,2(1)−1)=F′(x,1)=f″(x)
So now we only need to show that it is continuous to meet the requirements of a homotopy.

This map is indeed continuous, since at t=12 we have

F(x,2t)=f′(x)=F′(x,2t−1)
Therefore, by the Pasting Lemma, G s continuous on X×I
and is the required homotopy between f and f″.

Let’s also look at transitivity for path homotopies. Let:

F:I×I⟶X

F′:I×I⟶X
be path homotopies between f and f′, and f′ and f″ respectively, and define G nearly the same as before, only with variable s

G(s,t)={F(s,2t)F′(s,2t−1)t∈[0,12]t∈[12,1]

Since f and f′ are both path homotopies, we have that for all t∈I

F(0,t)=x0,F(1,t)=x1
and

F′(0,t)=x0,F′(1,t)=x1

So we have

G(0,t)={F(0,2t)F′(0,2t−1)t∈[0,12]t∈[12,1]=x0∀t∈[0,1]
and

G(1,t)={F(1,2t)F′(1,2t−1)t∈[0,12]t∈[12,1]=x1∀t∈[0,1]
Therefore, G is a path homotopy between f and f′.

Definition 1.4
We say that a set X is convex if every line segment connecting two points in X fully contained in X.

Theorem 1.5
Let f and g be any two maps of a space X into ℝ2. Then, f and g are homotopic and

F(x,t)=(1−t)f(x)+tg(x)
is a homotopy between them.

The proof of this comes straight from the definition of homotopy.

For an example, let

f:x↦(x,x2)

g:x↦(x,x5)
Below, we can see the homotopy

F(x,t)=(1−t)(x,x2)+t(x,x5)
transforming f into g as t goes from 0 to 1.

F in this case is called a straight line homotopy. The straight line homotopy will work for any convex subset A of ℝn and

f:X⟶A

g:X⟶A

Example 1.6
Let X=ℝ2−{0}. Aplanewithaholeinthemiddle The paths

f(s)=(cosπs,sinπs)

g(s)=(cosπs,2sinπs)
are path homotopic in X.

But if we let

h(s)=(cosπs,−sinπs)
f and h are not path homotopic. Why? Transforming f into h would require passing through (0,0)—where the hole is! Therefore the transformation cannot be continuous, which is required of a homotopy.

Definition 1.7
An equivalence class under homotopy is called a homotopy class. The homotopy class consisting of all maps homotopic to f is denoted [f].

Definition 1.8
Let f be a path in X from x0 to x1, and let g be a path in X from x1 to x2. We call these paths composable, since f(1)=g(0).


We define the product

h(s)=(f∗g)(s)={f(2s)g(2s−1)s∈[0,12]s∈[12,1]
This function is also sometimes called concatenation, and is well defined and continuous by the Pasting Lemma. This functions represents the path created by traveling along f from x0 to x1, and then along g from x1 to x2.

<

Theorem 1.9
Let f and f′ be two paths from x0 to x1 and let g and g′ be two paths from x1 to x2. If F is a path homotopy between between f and f′ and G is a path homotopy between g and g′, the product H=F∗G is a homotopy between f∗g and f′∗g′.



Proof: By the definition of the product we have that

H(s,t)=(F∗G)(s,t)={F(2s,t)G(2s−1,t)s∈[0,12]s∈[12,1]
This map is also well-defined and continuous, since at t=12 we have F,(1,t)=G(0,t)=x1 for all t.

Now, if we let t=0 we get

H(s,0)={F(2s,0)G(2s−1,0)s∈[0,12]s∈[12,1]

={f(2s)g(2s−1)s∈[0,12]s∈[12,1]=(f∗g)(s)
Similarly, for t=1 we have

H(s,1)={f′(2s)g′(2s−1)s∈[0,12]s∈[12,1]=(f′∗g′)(s)

Therefore, H is a homotopy between f∗g and f′∗g′. In other words, the product of the homotopies, is the homotopy of the products!


Corollary 1.10
For homotopy classes [f] and [g], we have the property

[f]∗[g]=[f∗g]

Theorem 1.11
The operation * has the following properties:

1. Associativity [f]∗([g]∗[h])=([f]∗[g])∗[h]
2. Identity [ex0]∗[f]=[f] and [f]∗[ex1]=[f] where$ea:x↦a$istheconstantfunctioninthesamespaceas$f$
3. Inverse Given the path f in X from x0 to x1, let f̃  be the path defined by
   f̃ (s)=f(1−s)
   f̃  is called the reverseof f and we have \\
   [f]∗[f̃ ]=[ex0] and [f̃ ]∗[f]=[ex1]

Before we can prove this, we need the following lemma:

Lemma 1.12

1. Let F be a path homotopy in X between the paths f and f′ and letk:X⟶Ybe any continuous map from the space X to the space Y.
   Then, k∘F is a path homotopy between the paths k∘f and k∘f′.
2. If f and g are paths in X with f(1)=g(0), then
   k∘(f∗g)=(k∘f)∗(k∘g)

Proof of Lemma:

     a Since F be a path homotopy, it is a continuous map

F:X×I⟶Ysuch that

F(s,0)=f(s),F(s,1)=f′(s)
and

F(0,t)=x0,F(1,t)=x1
Now if we compose k with F, we get a continuous map

k∘F:I×I⟶Y
such that

(k∘F)(s,0)=k(F(s,0))=k(f(s))=(k∘f)(s)

(k∘F)(s,1)=k(F(s,1))=k(f′(s))=(k∘f′)(s)
Additionally we have that

(k∘F)(0,t)=k(F(0,t))=k(x0)

(k∘F)(1,t)=k(F(1,t))=k(x1)
which, if x0 and x1 are the endpoints of f and f′, will be the endpoints of the paths k∘f and k∘f′.

Therefore k∘F is a path homotopy between k∘f and k∘f′.

     b Using the definition of the product:

k∘(f∗g)(s)=k∘{f(2s)g(2s−1)s∈[0,12]s∈[12,1]

={k∘f(2s)k∘g(2s−1)s∈[0,12]s∈[12,1]

=(k∘f)∗(k∘g)(s)

Now we are ready to prove the theorem.


Proof of Theorem 1.11: We will do these out of order since associativity is a bit trickier than the rest.

     II. The Identity Property: First we will show the first half of the statement, that [ex0]∗[f]=[f]. To do this, we will define two functions and show they are homotopic. Then, we will compose each with f, which by part a of our lemma will give us our result.
First, define the function

e0:I⟶Iby

e0(s)=0∀s∈[0,1]
and the function

i:I⟶Iby

i(s)=s∀s∈[0,1]
Now, take the product

(e0∗i)(s)={e0(2s)i(2s−1)s∈[0,12]s∈[12,1]

={02s−1s∈[0,12]s∈[12,1]
Since I is convex, by Theorem 1.5 we can define

G:I×I⟶Iby

G(s,t)=(1−t)i(s)+t(e0∗i)(s)
which is a homotopy of i and e0∗i.

Now we will use our lemma. We know that f∘G:I×I⟶I is continuous and:

(f∘G)(s,0)=f(G(s,0))=f(s)

(f∘G)(s,1)=f(G(s,1))=f(e0∗i)(s)=((f∘e0)∗(f∘i))(s)

={f∘e0(s)f∘i(2s−1)s∈[0,12]s∈[12,1]

={x0f(2s−1)s∈[0,12]s∈[12,1]

=(ex0∗f)(s)
This gives us f≃pex0∗f and our result

[f]=[ex0∗f]=[ex0]∗[f]
To show that [f]∗[ex1]=[f], we will use a similar method. Define

e1:I⟶Iby

e1(s)=1∀s∈[0,1]and keep i the same.
Now we have the product i∗e1:I⟶I given by

(i∗e1)(s)={2s1s∈[0,12]s∈[12,1]
Just like in the first part, we can define H:I×I⟶I to be a homotopy between i and i∗e1.

Therefore, f∘H is a homotopy between f=f∘i and (f∘i)∗(f∘e1)=f∗ex1, and we have our result

[f]≃p[f]∗[ex1]

     III. The Inverse Property: Using the same identity map i as before, we can define its reverse by

ĩ (s)=1−s
i∗ĩ  will be a path in I from 0 to 0, and therefore we have that

e0≃pi∗ĩ 
By our lemma, this gives us

f∘e0≃p(f∘i)∗(f∘ĩ )
where f∘e0=x0 and

(f∘i)∗(f∘ĩ )={f∘i(2s)f∘ĩ (2s−1)s∈[0,12]s∈[12,1]

={f(2s)f(−2s)s∈[0,12]s∈[12,1]

=f∗f̃ 
is a path that starts and ends at x0.

Therefore we have

f∗f̃ ≃pex0
and our intended result

[f]∗[f̃ ]=[ex0]
Through the same process we can show that

[f̃ ]∗[f]=[ex0]
With that, the time for associativity has come.

     I. The Associative Property To begin this proof, we have to establish the algebraic fact which goes as follows: Let [a,b] and [c,d] are two intervals in ℝ. There will always be a unique positive linear map

p:[a,b]⟶[c,d]

which transforms [a,b] into [c,d] and has the form y=mx+b. Positive here means that the map will always have a positive slope, which makes sense because we need to preserve order. For both intervals, as x increases, y increases.

For instance, the positive linear map from [1,2] to [−5,−1] is given by
p(x)=4x−9.

Now, let f,g,h be three composable functions. That is:

• f(0)=x0
• f(1)=x1=g(0)
• g(1)=x2=h(0)
• h(1)=x3

We can now consider the triple product of these three functions. \\ \\
First, consider the product f∗(g∗h), which is given by

(f∗(g∗h))(s)={f(2s)(g∗h)(2s−1)s∈[0,12]s∈[12,1]

=⎧⎩⎨⎪⎪f(2s)g(4s−2)h(4s−3)s∈[0,12]s∈[12,34]s∈[34,1]\\

Now what if we moved the parentheses? We would get

((f∗g)∗h)(s)={(f∗g)(2s)h(2s−1)s∈[0,12]s∈[12,1]

=⎧⎩⎨⎪⎪f(4s)g(4s−1)h(2s−1)s∈[0,14]s∈[14,12]s∈[12,1]

Our challenge is to prove that these are equal. To do this, we will look at the product as a series of compositions involving positive linear maps.


Looking at f∗(g∗h), the first component of the product maps the interval [0,12 to [0,1] and then plugs it into f. The second maps [12,34] to [0,1] and then plugs it into to g, and similarly for the third component.


Let’s define a path to represent these three linear maps. Let a and b be points such that 0<a<b<1. Define a path ka,b as follows:

• On the interval [0,a], ka,b equals the linear map of [0,a] to [0,1], plugged into f
• On the interval [a,b], it equals the linear map [a,b] to [0,1], plugged into g
• On the interval [b,1], it equals the linear map [b,1] to [0,1], plugged into h

The way we defined this path, it could represent either f∗(g∗h) or (f∗g)∗h.


We can show that if we replace a and b with two different points c and d where 0<c<d<1, then ka,b≃Pkc,d.


Let p:I⟶I be a map where p(a)=c and p(b)=d.



p is made up of the positive linear maps that map [0,a], [a,b], and [b,1] to [0,c], [c,d], and [d,1] respectively, and we have

ka,b∘p=kc,d
Since p turns a into c and b into d.

Now, since p maps I to I and so does the identity map,
i:I⟶I, there must exist a path homotopy P between them. By Lemma 1.12, this means that ka,b∘P is a path homotopy between ka,b and kc,d.

Hence, f∗(g∗h) and (f∗g)∗h are in the same homotopy class. This gives us our result

[f]∗([g]∗[h])=([f]∗[g])∗[h]
In fact, this extends to associativity for any finite product of paths!

Theorem 1.13
Let f be a path in X and let

0=a0<a1<⋯<an=1
be a partition of the interval [0,1] and let

fi:I⟶Xi=1,2,…,n
be the path that represents the positive linear map of I onto [ai−1,ai] plugged into fi.

Then,

[f]=[f1]∗[f2]∗⋯∗[fn]

Proof: Consider the set of paths

pi:I⟶Igiven by

pi(s)=(1−s)ai−1+sai0≤s≤1,i=1,2,…,n
Since I is convex, all of these paths are fully contained in I. All of these paths together are essentially connecting the dots a0,a1,…,an.

Now define P:I⟶I by

p=p1∗p2∗⋯∗pn
Since all of these together create a path from 0 to 1, we have that

[p]=[p1]∗[p2]∗⋯∗[pn]=[i]
where i is the identity map in I,
and

p(0)=p1(0)=0=i(0)

p(1)=pn(1)=1=i(1)
Therefore, we can say that there is a path homotopy F:I×I⟶I between p and i which satisfies

F(s,0)=p(s)F(s,1)=i(s)

F(0,t)=0F(1,t)=1
Finally, we can compose each with f and get

f∘i≃(f∘p1)∗(f∘p2)∗⋯∗(f∘pn)
which gives us the intended result

[f]=[f1]∗[f2]∗⋯∗[fn]

Problem 1
Show that if

h,h′:X⟶Yare homotopic and

h,k′:Y⟶Zare also homotopic, then k∘h and k′∘h′
are homotopic.

Proof: Since we know h≃h′, there exists a homotopy

H:X×I⟶Y

H(x,0)=h(x),H(x,1)=h′(x)
Similarly, since k≃k′ we have the homotopy

K:Y×I⟶Z

K(y,0)=h(y),K(y,1)=h′(y)
Now, to define a homotopy between k∘h and k′∘h′, we need to build a composition map. So define a new function

Φ:X×I⟶Y×I
given by

Φ(x,t)=(H(x,t),t)
and define

F:X×I⟶Z
to be a continuous map.
Now we can build a complete diagram to connect the three spaces:

This shows us visually that F=K∘Φ. So we have

F(x,0)=(K∘Φ)(x,0)=K(H(x,0),0)

=K(h(x),0)=(k∘h)(x)
and

F(x,1)=(K∘Φ)(x,1)=K(H(x,1),1)

=K(h′(x),1)=(k′∘h′)(x)
Therefore F is the required homotopy which gives us

k∘h≃k′∘h′

Problem 2
Given the spaces X and Y, let [X,Y] denote the set of homotopy classes of maps of X into Y.

1. Let I=[0,1]. Show that for any X, the set [X,I] has only one element.
2. A subset A of a space X is called path connected if, for every pair of points a,b∈A, there exists a continuous mapφ:[0,1]⟶Asuch that φ(0)=a and φ(1)=b.
   Show that if Y is path connected, the set [I,Y] has a single element. /li>

Proof: \\
     a To show that the set of homotopy classes has only one element, we need to show that if f and g are two maps between X and I, it must be true that f≃g. So define

f,g:X⟶I
If I is convex, we can define the straight line homotopy between them

F(x,t)=(1−t)g(x)+tf(x)
We’ve been treating I as convex throughout this section, but it doesn’t hurt to prove it here. To show that a set is convex, we need to show that for any two points a,b∈I, the straight line connecting them is also in I.

To do this, let a≤b be points in I. The line between them is given the same way that the homotopy is, for t∈[0,1] we have that

c(t)=(1−t)a+tb
connects a and b, and furthermore, given how we restrict t every point on that line is in I. Hence, I is convex and for any f,g that map X to I

F(x,t)=(1−t)g(x)+tf(x)
must be a homotopy between them.

Therefore [X,Y] contains only one element.

     b For this problem, we define the map

ea:I⟶Y
by the constant function

ea(s)=a∀s∈[0,1]
where a∈Y is a fixed point. To prove that [I,Y] has only one element, we will prove that any other continuous map

g:I⟶Ymust be homotopic to ea.

We can show that g≃eg(0) by defining the homotopy

G:I×I⟶Yby

G(x,t)=g((1−t)x)

which gives us the required

G(x,0)=g(0),G(x,1)=g(x)

However, this is not enough to say [I,Y]={[eg(0)]} because the value of eg(0) depends on g and we want this to work for any continuous function from I to Y.
However, if we can show that ea≃eb for all a,b∈Y, we can use g≃eg(0) to show that g≃ea.

Since Y is path connected, for any a,b∈Y, there exists a path

φ:I⟶Y
such that

φ(0)=a,φ(1)=b
We can use this path to give us the required homotopy between ea and eb. Consider the map

E:I×I⟶Y
defined by

E(s,t)=φ(t)
We have

E(s,0)=φ(0)=a=ea(s)

E(s,1)=φ(1)=b=eb(s)
Therefore ea≃eb.

So, for any path g:I⟶Y we know that
and now we know g≃eg(0), and we now know that

g≃eg(0) and eg(0)≃ea
which together give us

g≃ea⇒[g]=[ea]
Hence, the set [I,Y] has a single element.

Problem 3
A space X is said to be contractibleif the identity map is null homotopic. Show that:

1. I and ℝ are contractible.
2. A contractible space is path connected.
3. If Y is contractible, then for any X, the set [X,Y] has a single element.
4. If X is contractible and Y is path-connected, the [X,Y] has a single element.

Proof:

     a Now this certainly feels like it should be true, looking at a picture for the identity map in ℝ.




Let’s start with the proof for I. Let iI:I⟶I denote the identity function in I. We only need to prove that the identity map is equal to some constant map, so we’ll use the easiest one:

0:I⟶Iwhere

0(s)=0∀s∈IConsider

F:I×I⟶I
where

F(s,t)=stThis gives us

F(s,0)=0⋅s=0(s)

F(x,1)=1⋅s=iI(s)
Therefore, iI≃0.

Now, for ℝ, we redefine our 0 function as

0:ℝ⟶ℝ

0(s)=0∀s∈ℝand adjust our F to be

F:ℝ×I⟶ℝ
where

F(s,t)=st
And just as before we have

F(s,0)=0⋅s=0(s)

F(x,1)=1⋅s=iℝ(s)
which gives us our result of iℝ≃0.

     b Let X be a contractible space. We need to show that for all a,b∈X, there exists a path φ:[0,1]⟶X which connects them.
Let iX:X⟶X be the identity function in X and define

ea:I⟶X
by

ea(s)=a∀s∈[0,1]where a∈X is fixed.
Since X is contractible, we have iX≃ea. Therefore there exists a continuous map

F:X×I⟶Xsuch that

F(x,0)=ea(x)=a,F(x,1)=iX(x)=x
This is almost enough. We just need adjust our function so it can map a to any point b∈X.

To do this, define

φ:I⟶Xby

φ(t)=F(b,t)
This gives us exactly what we want

φ(0)=F(b,0)=ea(b)=a

φ(1)=F(b,1)=iX(b)=b
Since we can choose any a,b∈X here that we’d like, we have our result that X is path connected.

%3c
     c Let

f:X⟶Ybe a continuous map. Just like in Problem 2b, we will get our result by showing that f is homotopic to a constant map.
Since Y is contractible, there exists a continuous map F:Y×I⟶Y such that

F(y,0)=ea(x)=a

F(y,1)=iY(y)=y
To use this knowledge to say anything about maps from X⟶Y, we need do to some function composition. So define the function

Φ:X×I⟶Y×Iby

Φ(x,t)=(f(x),t)We now have the composition diagram

which shows us that

G(x,t)=(F∘Φ)(x,t)=F(f(x),t)
Therefore, if F and Φ are continuous, G must be continuous and

G(x,0)=F(f(x),0)=ea(x)=a

G(x,1)=F(f(x),1)=f(x)

Hence, f≃a and the set [X,Y] has only one element.

     d Let X be a contractible space. Thus we know, for any p∈X that there exists a continuous H:X×I⟶X such that

H(x,0)=IX(x)=x

H(x,1)=p
Now, pick a point q∈Y and define the constant map

fq:X⟶Y
by

fq(x)=q∀x∈X
Once again, we want to show that for any continuous function,

g:X⟶Y
g must be homotopic to fq.

Given q and g(p) are both points in Y, by path-connectedness there must exist a map

φ:I⟶Y
such that

φ(0)=g(p),φ(1)=q
Now define the function

Φ(x,t)={g(H(x,2t))φ(2t−1)t∈[0,12]s∈[12,1]
Φ is continuous an well defined at t=12 since g(H(x,1))=g(p)=φ(0), and since

Φ(x,0)=g(H,x,0))=g(x)
and

Φ(x,1)=φ(1)=fq(x)
we have that g≃fq. Therefore, [X,Y] has one element.