Notation 2.1.1. \(I=[0,1]\).
Throughout this course \(I\) will denote the interval \([0,1]\text{.}\)
Section 2.1 Homotopy
Definition 2.1.2. Homotopy of maps.
Let \(f,g\colon X\rightarrow Y\) be continuous maps. A homotopy from \(f\) to \(g\) is a continuous map
\begin{equation*}
F\colon X\times I\rightarrow Y
\end{equation*}
satisfying
\begin{align*}
F(x,0) \amp =f(x) \amp F(x,1)\amp= g(x)
\end{align*}
for all \(x\in X\text{.}\) We say \(f\) is homotopic to \(g\) in this case, written \(f\simeq g\text{.}\)
Remark 2.1.3.
Given a homotopy \(F\colon X\times [0,1]\rightarrow Y\) from \(f\) to \(g\text{,}\) for each \(t\in I\) we get a function \(f_t\colon X\rightarrow Y\) defined as \(f_t(x)=F(x,t)\text{.}\) In this way we can think of a homotopy \(F\) as a family of functions \(\{f_t\}_{t\in I}\) that “varies continuously with \(t\)” (a shorthand way of saying that \(F\) is continuous), and which “continuously deforms” the function \(f=f_0\) to the function \(g=f_1\) as \(t\) ranges over \(I\text{.}\)
Example 2.1.4. Homotopic functions to \(\R^n\).
A subset \(A\subseteq \R^n\) is convex if for any elements \(\boldx, \boldy\in A\) and any \(t\in I\text{,}\) we have \((1-t)\boldx+t\boldy\in A\text{:}\) i.e., given any pair of points in \(A\text{,}\) the line segment between them lies within \(A\text{.}\)
If \(A\subseteq \R^n\) is convex, then given any continuous functions \(f,g\colon X\rightarrow A\text{,}\) the function
\begin{align*}
F\colon X\times I \amp \rightarrow A\\
(x,t) \amp \mapsto (1-t)f(x)+tg(x)
\end{align*}
defines a homotopy from \(f\) to \(g\text{.}\) Thus any two continuous functions from a space \(X\) into \(A\) are homotopic.
Definition 2.1.5. Nullhomotopic map.
A continuous map \(f\colon X\rightarrow Y\) is nullhomotopic if it is homotopic to a constant function from \(X\) to \(Y\text{.}\)
Definition 2.1.6. Path homotopy.
Given elements \(x_0, x_1\) in the topological space \(X\text{,}\) we denoted by \(P(X; x_0,x_1)\) the set of all paths \(f\colon [0,1]\rightarrow X\) with initial point \(x_0=f(0)\) and terminal point \(x_1=f(1)\text{.}\)
Given \(f,g\in P(X; x_0, x_1)\text{,}\) a path homotopy from \(f\) to \(g\) is a continuous function \(F\colon I\times I\rightarrow X\) satisfying
\begin{align*}
F(s,0) \amp = f(s) \amp F(s,1)\amp = g(s) \text{ for all } s\in I\\
F(0,t) \amp =x_0 \amp F(1,t)\amp =x_1 \text{ for all } t\in I\text{.}
\end{align*}
In other words, \(F\) is a homotopy \(\{f_t\}_{t\in I}\) from \(f=f_0\) to \(g=f_1\) that further satisfies \(f_t(0)=x_0\) and \(f_t(1)=x_1\) for all \(t\in I\text{.}\) We say that \(f\) is path homotopic to \(g\) in this case, written \(f\simeq_p g\text{.}\)
Example 2.1.7. Homotopic paths in \(\R^n\).
Let \(A\) be a convex subset of \(\R^n\text{,}\) and let \(\boldx_0, \boldx_1\in A\text{.}\) Given two paths \(f,g\in P(A; \boldx_0, \boldx_1)\text{,}\) the straight-line homotopy \(F(s,t)=(1-t)f(s)+tg(s)\) is easily seen to be a path homotopy:
\begin{align*}
F(0,t)\amp =(1-t)f(0)+tg(0)=(1-t)\boldx_0+t\boldx_0=\boldx_0 \\
F(1,t)\amp =(1-t)f(1)+tg(1)=(1-t)\boldx_1+t\boldx_1=\boldx_1 \text{.}
\end{align*}
Thus any two paths to \(A\) between any two points \(\boldx_0, \boldx_1\in A\) are path homotopic.
Example 2.1.8. Paths with non-convex codomain.
Consider the the nonconvex set \(A=\R^2-\{(0,0)\}\) and the paths
\begin{align*}
f\colon [0,1] \amp \rightarrow A \amp g\colon [0,1]\amp \rightarrow A\\
s \amp \mapsto (\cos (2\pi s), \sin (2\pi s))\amp s\amp \mapsto (1,0)\text{.}
\end{align*}
Both \(f, g\) are paths that begin and end at \((1,0)\text{;}\) it is intuitively clear that \(f\) cannot be continuously deformed to \(g\) (while staying tethered at \((1,0)\)) without passing through the origin \((0,0)\) at some point. In other words, the two paths are not path homotopic. This is surprisingly difficult to prove rigorously; we will get to this in the coming sections. On the other hand \(f\) and \(g\) are in fact homotopic: consider \(F(s,t)=\left(\cos(2\pi(1-t)s),\sin(2\pi(1-t)s)\right)\)
Interestingly, the situation is different if we take \(A=\R^3-\{(0,0,0)\}\) and
\begin{align*}
f\colon [0,1] \amp \rightarrow A \amp g\colon [0,1]\amp \rightarrow A\\
s \amp \mapsto (\cos (2\pi s), \sin (2\pi s),0)\amp s\amp \mapsto (1,0,0)\text{.}
\end{align*}
Intuitively, it seems we should be able to “lift” the graph of \(f\) up over the origin, then shrink it down to the point \((1,0,0)\text{.}\) Here is one explicit realization of this idea. Consider the family of curves \(C_t\) lying on the upper hemisphere of \(S^2: x^2+y^2+z^2=1\text{,}\) whose projection onto the \(xy\)-plane is the circle \((x-t)^2+y^2=(1-t)^2\text{.}\) Such a curve has parametrization
\begin{equation*}
C_t(s)=\left(t+(1-t)\cos 2\pi s, (1-t)\sin 2\pi s, \sqrt{1-(t+(1-t)\cos 2\pi s)^2+(1-t)\sin 2\pi s)^2}\right)\text{,}
\end{equation*}
giving rise to a path homotopy \(H(s,t)=C_t(s)\) from \(f\) to \(g\text{.}\) The Sage cell below provides a nice visualization of this family of curves.
Theorem 2.1.9. Homotopy equivalence relation.
Let \(X\) and \(Y\) be topological spaces. Fix \(x_0, x_1\in X\text{.}\)
- \(\simeq\) is an equivalence relation on the set of all continuous functions from \(X\) to \(Y\text{.}\)
- \(\simeq_p\) is an equivalence relation on \(P(X; x_0,x_1)\text{.}\)
Proof.
We give a proof for (a). It is then easy to check that the same homotopies provided in this proof are path homotopies (i.e., satisfy \(F(0,t)=x_0, F(1,t)=x_1\)) in the context of paths in \(P(X; x_0,x_1)\text{.}\)
Reflexive.
Given a continuous function \(f\colon X\rightarrow Y\text{,}\) the map defined as \(F(x,t)=f(x)\) for all \((x,t)\in X\times I\) is clearly a homotopy from \(f\) to \(f\text{.}\)
Symmetric.
Let \(f,g\) be continuous maps from \(X\) to \(Y\text{,}\) and assume \(f\simeq g\text{.}\) By definition this means there is a homotopy \(F\colon X\rightarrow I\rightarrow Y\) from \(f\) to \(g\text{,}\) so that in particular \(F(x,0)=f(x), F(x,1)=g(x)\) for all \(x\in X\text{.}\) Define \(G\colon X\rightarrow I\rightarrow Y\) as \(G(x,t)=F(x,1-t)\text{.}\) It is clear that \(G\) is continuous, and we have
\begin{align*}
G(x,0) \amp = F(x,1)=g(x) \amp G(x,1)\amp =F(x,0)=f(x)\text{.}
\end{align*}
Thus \(G\) is a homotopy from \(g\) to \(f\text{,}\) showing \(g\simeq f\text{.}\)
Transitive.
Let \(f,g,h\) be continuous maps from \(X\) to \(Y\text{.}\) We assume \(f\simeq g\) and \(g\simeq h\text{.}\) By definition there is a homotopy \(F\) from \(f\) to \(g\text{,}\) and a homotopy \(G\) from \(g\) to \(h\text{.}\) Define
\begin{align*}
H(x,t) \amp =\begin{cases}
F(x,2t) \amp \text{if } t\in [0,1/2]\\
G(x,2t) \amp \text{if } t\in [1/2, 1]
\end{cases}\text{.}
\end{align*}
The functions \(F(x,2t)\) and \(G(x,2t-1)\) are continuous on the closed sets \(X\times [0,1/2]\) and \(X\times [1/2, 1]\text{,}\) respectively, and agree on the intersection \(X\times \{1/2\}\) since \(F(x,2(1/2))=F(x,1)=g(x)\) and \(G(x,2(1/2)-1)=G(x,0)=g(x)\) for all \(x\in X\text{,}\) by definition of homotopy. By the closed pasting lemma, we conclude that \(H\) is well-defined and continuous. Lastly, we have
\begin{align*}
H(x,0) \amp =F(x,0)=f(x)\\
H(x,1) \amp =G(x,1)=g(x)
\end{align*}
for all \(x\in X\text{.}\)
Notation 2.1.10. Homotopy equivalence.
For topological spaces \(X, Y\text{,}\) let \(C(X,Y)\) be the set of all continuous functions. For a function \(f\in C(X,Y)\) we let \([f]\) denote its equivalence class with respect to \(\simeq\text{.}\) The set \(C(X,Y)/\simeq\) of all equivalence classes is denoted \([X,Y]\text{.}\)
For \(f\in P(X; x_0,x_1)\text{,}\) we will also use \([f]\) to denote its equivalence class with respect to \(\simeq_p\text{.}\)
Definition 2.1.11. Path product.
Let \(X\) be a topological space, and let \(x_0,x_1,x_2\in X\text{.}\) We define a path product operation
\begin{align*}
P(X;x_0,x_1)\times P(X; x_1,x_2) \amp \rightarrow P(X;x_0, x_2)\\
(f,g) \amp \mapsto f*g
\end{align*}
as follows:
\begin{equation*}
f*g(s)=\begin{cases}
f(2s)\amp \text{if } s\in [0,1/2]\\
g(2s-1)\amp \text{if } s\in [1/2,1]
\end{cases}\text{.}
\end{equation*}
The function \(f*g\) defined is a path from \(x_0\) to \(x_2\text{.}\) Furthermore, if \(f\simeq_p f'\) and \(g\simeq_p g'\text{,}\) then \(f*g\simeq f'*g'\text{.}\) In other words, we get a well-defined operation on homotopy classes defined as
\begin{equation*}
[f]*[g]=[f*g]\text{.}
\end{equation*}
Proof.
It is easy to see, using the closed pasting lemma, that \(f*g\) is well-defined and continuous. We show that it is well-defined on homotopy classes, as claimed.
Suppose that \(f\simeq_p f'\) as witnessed by the path homotopy \(F\text{,}\) and that \(g\simeq g'\) as witnessed by the path homotopy \(G\text{.}\) Define
\begin{align*}
H(s,t) \amp =
\begin{cases}
F(2s,t)\amp \text{if } s\in [0,1/2]\\
G(2s-1,t)\amp \text{if } s\in [1/2, 1]
\end{cases}\text{.}
\end{align*}
Note first that \(F(2(1/2),t)=F(1,t)=x_1\) and \(G(2(1/2)-1,t)=G(0,t)=x_1\text{,}\) since \(F\) and \(G\) are path homotopies. Since \(F(2s,t)\) is continuous on \([0,1/2]\times I\) and \(G(2s-1,t)\) is continuous on \([1/2,1]\times I\text{,}\) and since they agree on the intersection of these sets, the closed pasting lemma implies \(H\) is a well-defined continuous function.
Next, we have
\begin{align*}
H(s,0) \amp =\begin{cases}
F(2s,0)\amp \text{if } s\in [0,1/2]\\
G(2s-1,0)\amp \text{if } s\in [1/2, 1]\\
\amp =\begin{cases}
f(2s)\amp \text{if } s\in [0,1/2]\\
g(2s-1)\amp \text{if } s\in [1/2, 1] \\
\amp =f*g(s)\\
H(s,1) \amp =\begin{cases}
F(2s,1)\amp \text{if } s\in [0,1/2]\\
G(2s-1,1)\amp \text{if } s\in [1/2, 1]\\
\amp =\begin{cases}
f'(2s)\amp \text{if } s\in [0,1/2]\\
g'(2s-1)\amp \text{if } s\in [1/2, 1] \\
\amp =f'*g'(s)\text{,}
\end{align*}
showing that \(H\) is a homotopy from \(f*g\) to \(f'*g'\text{.}\)
Lastly, we have
\begin{align*}
H(0,t) \amp =F(0,t)=x_0\\
H(1,t) \amp =G(2(1)-1,t)=G(1,t)=x_2
\end{align*}
for all \(t\in I\text{,}\) showing that \(H\) is a path homotopy.
Theorem 2.1.12. Path product properties.
Let \(X\) be a topological space. For any element \(x\in X\text{,}\) we let \(e_x\colon I\rightarrow X\) denote the constant function \(e_x(s)=x\) for all \(s\in I\text{.}\)
-
Associativity.If \(f\in P(X; x_0,x_1), g\in P(X; x_1,x_2), h\in P(X; x_2,x_3)\text{,}\) then\begin{equation*} [f]*([g]*[h])=([f]*[g])*[h]\text{.} \end{equation*}
-
Identities.Given \(f\in P(X; x_0, x_1)\text{,}\) we have\begin{equation*} [e_{x_0}]*[f]=[f]*[e_{x_1}]=[f]. \end{equation*}
-
Inverses.Given \(f\in P(X; x_0,x_1)\text{,}\) let \(\overline{f}\in P(X; x_1,x_0)\) be its reverse: i.e., \(\overline{f}(s)=f(1-s)\text{.}\) We have\begin{align*} [f]*[\overline{f}] \amp =[e_{x_0}] \amp [\overline{f}]*[f]=[e_{x_1}]\text{.} \end{align*}
Proof.
Associativity. Let \(p=f*(g*h)\) and \(q=(f*g)*h\text{.}\) After unpacking the definitions for iterated path products, we see that
\begin{align*}
p(s) \amp = \begin{cases}
f(2s)\amp \text{if } s\in [0,1/2]\\
g(4s-2)\amp \text{if } s\in [1/2,3/4]\\
h(4s-3)\amp \text{if } s\in [3/4,1]
\end{cases}\\
q(s) \amp = \begin{cases}
f(4s)\amp \text{if } s\in [0,1/4]\\
g(4s-1)\amp \text{if } s\in [1/4,1/2]\\
h(2s-1)\amp \text{if } s\in [1/2,1]
\end{cases}
\end{align*}
Define \(\phi\colon I\rightarrow I\) as
\begin{align*}
\phi(s) \amp =\begin{cases}
\frac{1}{2}s \amp \text{if } s\in [0,1/4]\\
s-1/4 \amp \text{if } s\in [1/4,3/4]\\
2s-1 \amp \text{if } s\in [3/4, 1]
\end{cases}\text{.}
\end{align*}
The pasting lemma implies \(\phi\) is continuous, and by definition we have \(\phi(0)=0\) and \(\phi(1)\text{.}\) A computation shows that \(f*(g*h)(s)=(f*g)*h(\phi (s))\text{.}\) In other words, \(p=q\circ \phi\text{.}\) By Lemma 2.1.13 we conclude that \([f*(g*h)]= [(f*g)*h]\) and thus that
\begin{align*}
[f]*([g]*[h]) \amp =[f]*[g*h] \amp (\text{def.}) \\
\amp = [f*(g*h)] \amp (\text{def.})\\
\amp = [(f*g)*h]\amp (\text{by above})\\
\amp = [f*g]*[h] \amp (\text{def.)\\
\amp = ([f]*[g])*[h]\amp (\text{def.})\text{.}
\end{align*}
Identities. Using the definition of path product, we have
\begin{align*}
e_{x_0}*f(s) \amp =
\begin{cases}
x_0\amp \text{if } s\in [0,1/2]\\
f(2s-1) \amp \text{if } s\in [1/2,1]
\end{cases}
\end{align*}
Define \(\phi\colon I\rightarrow I\) as
\begin{align*}
\phi(s) \amp =
\begin{cases}
x_0 \amp \text{if } s\in [0,1/2]
2s-1 \amp \text{if } s\in [1/2/1]
\end{cases}
\end{align*}
Inverses. First compute
\begin{align*}
f*\overline{f}(s) \amp =
\begin{cases}
f(2s) \amp \text{if } s\in [0,1/2]\\
f(2(1-s)) \amp \text{if } s\in [1/2, 1]
\end{cases}\\
[\overline{f}]*[f]=[e_{x_1}]
\end{align*}
Lemma 2.1.13. Reparametrization of path.
Let \(f\in P(X; x_0, x_1)\text{.}\) If \(\phi\colon I\rightarrow I\) is a continuous map satisfying \(\phi(0)=0\) and \(\phi(1)=1\text{,}\) then \(f\simeq_p f\circ \phi\text{.}\) In other words, a “reparametrized” path is homotopic to the original path.
Proof.
Homework exercise.
Theorem 2.1.14. Partitions of \(I\).
Let \(f\in P(X; x_0,x_1)\text{,}\) let
\begin{equation*}
0=a_0< a_1 < a_2 < \dots < a_n=1
\end{equation*}
be a subdivision of \(I\) into \(n\) subintervals \(I_k=[a_{k-1}, a_k]\text{,}\) \(1\leq k\leq n\text{,}\) and let \(f_k=f\circ\phi_k\text{,}\) where \(\phi_k\colon I\rightarrow [a_{k-1}, a_k]\) is any continuous map satisfying \(\phi_k(0)=a_{k-1}\) and \(\phi_k(1)=a_k\text{.}\) We have
\begin{equation*}
[f]=[f_1]*[f_2]*\cdots *[f_n]\text{.}
\end{equation*}
Proof.
Homework exercise.
