Proving Munkres Theorem 51.3

Question

I am trying to prove Theorem 51.3 from Munkres, that is:

Let $f$ be a path in $X$ and let $a_0, \dotsc, a_n$ be numbers such that $0 = a_0 < a_1 < \dotsb < a_n = 1$. Let $f_i\colon I \to X$ be the path that equals the positive linear map of $I$ onto $[a_{i-1}, a_i]$ followed by $f$. Then \begin{equation*} [f] = [f_1] * \dotsb * [f_n]. \end{equation*}

I have considered a proof by induction. I have done the base case which follows directly, but I'm struggling with the induction step. Since the homotopy of paths is an equivalence relation, it would be equivalent to proving that $f$ and $f_{1} * f_{2} \dots * f_{n+1}$ are path homotopic given that $[f] = [f_1] * \dotsb * [f_n]$.

Perhaps my understanding of the problem is a bit skewed, I have tried drawing a picture to make sense of how to prove the theorem by inducition but that hasn't helped me. Any help is appreciated.

Answer 1

The expression $[f_1] * \ldots * [f_n]$ only makes sense because Munkres has proved in Theorem 51.2 that $*$ is associative. The following considerations will prove this again.

Let us consider arbitrary paths $f_1, \ldots, f_n : I \to X$ such that $f_{i-1}(1) = f_i(0)$ for $i=2,\ldots, n$. For any partition $\mathbf a = (a_0,\ldots, a_n)$ of $I$ with $0 = a_0 < a_1 < \ldots < a_n = 1$ we define $$f_\mathbf a : I \to X, f_{\mathbf a}(t) = f_i \Big( \frac{t-a_{i-1}}{a_i -a_{i-1}} \Big) \text{ for } t \in [a_{i-1},a_i] .$$ This is a well-defined continuous map.

Theorem. The path-homotopy equivalence class $[f_\mathbf a]$ does not depend on the choice of the partition $ \mathbf a$.

Let us start with a simple lemma.

Lemma. Each map $u : I \to I$ with $u(0) = 0$ and $u(1) = 1$ is homotopic to the identity map rel. $\{0,1\}$.

Proof. Define $H : I \times I \to I, H(s,t) = (1-t) u(s) + ts$. This is a homotopy from $u$ to $id$ which keeps the points $0, 1$ fixed. $\phantom{xx} \square$

We now prove the Theorem.

Let $\mathbf b = (b_0,\ldots, b_n)$ be another partition of $I$.

Define $u : I \to I$ by $$u(t) = b_{i-1} + \frac{b_i- b_{i-1}}{a_i - a_{i-1}}(t- a_{i-1}) \text { for } t \in [a_{i-1},a_i] .$$ This is a well-defined continuous map which maps $[a_{i-1},a_i]$ affinely onto $[b_{i-1},b_i]$. We have $$f_\mathbf a = f_\mathbf b \circ u.$$ In fact, for $t \in [a_{i-1},a_i]$ we get $$(f_\mathbf b \circ u)(t) = f_\mathbf b \left(b_{i-1} + \frac{b_i- b_{i-1}}{a_i - a_{i-1}}(t- a_{i-1})\right)= f_i\left(\frac{b_{i-1} + \frac{b_i- b_{i-1}}{a_i - a_{i-1}}(t- a_{i-1}) - b_{i-1}}{b_i - b_{i-1}}\right) \\ = f_i\left(\frac{t- a_{i-1}}{a_i - a_{i-1}} \right) = f_\mathbf a (t) .$$

Since $u \simeq id$ rel. $\{0,1\}$, we see that $$f_\mathbf a = f_\mathbf b \circ u \simeq f_\mathbf b \circ id = f_\mathbf b \text{ rel. } \{0,1\} . \phantom{xx} \square $$ Corollary. The product $*$ is associative on path-homotopy equivalence classes.

Proof. Consider $f_1, f_2, f_3: I \to X$ such that $f_1(1) = f_2(0)$ and $f_2(1) = f_3(0)$. With $\mathbf a =(0,\frac 1 2, \frac 3 4, 1)$ and $\mathbf b =(0,\frac 1 4, \frac 1 2, 1)$ we have $f_\mathbf a = f_1 * (f_2 * f_3)$ and $f_\mathbf b = (f_1 * f_2) * f_3$. We conclude $$[f_1] *([f_2] * [f_3]) = [f_1 * (f_2 * f_3)] = [f_\mathbf a] = [f_\mathbf b] = [(f_1 * f_2) * f_3] = ([f_1] * [f_2]) * [f_3] .\phantom{x} \square$$

Given $f_1,\ldots, f_n$ as above, a representative $F$ of $[f_1] * \ldots * [f_n]$ can be constructed by successively computing $F_1 = f_1$, $F_{i+1} = F_i * f_{i+1}$. Then $F = F_n$ is the desired path. By construction (formally we have to use induction to prove it), we get $F = f_\mathbf p$ with $\mathbf p = (p_0,\ldots,p_n)$, where $p_0 = 0$ and $p_i = 2^{i-n}$ for $i = 1,\ldots, n$. For $n = 2$ this means $\mathbf p = (0, \frac 1 2,1)$, for $n = 3$ it means $\mathbf p = (0, \frac 1 4, \frac 1 2,1)$, etc.

Corollary. Let $\mathbf a = (a_0,\ldots,a_n)$ be any partition of $I$. Then $[f_1] * \ldots * [f_n] = [f_\mathbf a]$.

Proof. $[f_1] * \ldots * [f_n] = [f_\mathbf p] = [f_\mathbf a]$. $\phantom{xx} \square$

Let us now prove Munkres's Theorem 51.3.

Munkres takes the "positive linear" map of $I$ onto $[a_{i-1},a_i]$ which is $$\phi_i : I \to [a_{i-1},a_i], \phi_i(t) = a_{i-1} + t(a_i - a_{i-1}) $$ and then defines $$f_i = f \mid_{[a_{i-1},a_i]} \circ \phi_i .$$

We have proved above that $$[f_1] * \ldots * [f_n] = [f_\mathbf a] .$$ On $[a_{i-1},a_i]$ we have $$f_\mathbf a(t) = f_i\Big(\frac{t-a_{i-1}}{a_i -a_{i-1}}\Big) = f\Big(\phi_i\Big(\frac{t-a_{i-1}}{a_i -a_{i-1}}\Big)\Big) = f\Big(a_{i-1} + \frac{t-a_{i-1}}{a_i -a_{i-1}}(a_i - a_{i-1} \Big) = f(t) .$$ This shows that $f_\mathbf a = f$, i.e. $$[f_1] * \ldots * [f_n] = [f] .$$