Munkres §51: Homotopy of Paths


A basic problem of topology is determining whether two spaces are homeomorphic. If so, then a continuous mapping between them with continuous inverse exists. Otherwise, if one can find a topological property that holds for one but not the other, then they cannot be homeomorphic:

  • $\left(0,1\right)$ and $\left[0,1\right]$ are not homeomorphic because the latter is compact and the former is not.
  • $\mathbb R$ and the long line are not homeomorphic because the former has a countable basis and the latter does not.
  • $\mathbb R$ and $\mathbb R^2$ are not homeomorphic because deleting a point from the latter leaves a connected space while doing so from the former does not.
However, distinguishing between $\mathbb R^2$ and $\mathbb R^3$ is more involved: deleting a point from the latter leaves a simply connected space while doing so from the former does not.

To distinguish between more spaces, a more general tool, their fundamental groups, can be used: two spaces are homeomorphic if their fundamental groups are isomorphic. For example, if a space is simply connected, then its fundamental group is trivial.

These writings build the tools necessary to talk about the fundamental group.

Claim: If $h,h':X\to Y$; $k,k':Y\to Z$; $h\simeq h'$; and $k\simeq k'$, then $k\circ h\simeq k'\circ h'$.

Proof: $(x,t)\mapsto G(F(x,t),t)$ is a homotopy.
$$\tag*{$\blacksquare$}$$

Claim: $[X,I]$ is a singleton.

Proof: $(x,t)\mapsto(1-t)f(x)+tg(x)$ is a homotopy.
$$\tag*{$\blacksquare$}$$
Claim: If $Y$ is path-connected, then $[I,Y]$ is a singleton.

Proof: If $h$ is a path from $f(1)$ to $g(0)$, then $f,g\in[I,Y]$ are closed subpaths of and thus homotopic to $f*h*g$.
$$\tag*{$\blacksquare$}$$
Claim: $I$ and $\mathbb R$ are contractible.

Proof: $(x,t)\mapsto (1-t)x$ is a homotopy, i.e., $\text{id}_I$ and $\text{id}_\mathbb R$ are nulhomotopic to $0$.
$$\tag*{$\blacksquare$}$$
Claim: If $X$ is contractible, then $X$ is path-connected.

Proof: $F(x,\cdot)$ is a path from $x$ to $x_0$, where $F$ is a homotopy between $\text{id}_X$ and $x_0$.
$$\tag*{$\blacksquare$}$$
Claim: If $Y$ is contractible, then $[X,Y]$ is a singleton.

Proof: $\text{id}_Y\circ f=f$ is homotopic to $y_0\circ f=y_0$ since $\text{id}_Y$ is homotopic to $y_0$.
$$\tag*{$\blacksquare$}$$
Claim: If $X$ is contractible and $Y$ is path-connected, then $[X,Y]$ is a singleton.

Proof: $f\circ\text{id}_X=f$ is homotopic to $f\circ x_0=f(x_0)$ since $\text{id}_X$ is homotopic to $x_0$ and all constant maps are nulhomotopic since $Y$ is path-connected.
$$\tag*{$\blacksquare$}$$

Two Pathological Examples in Functional Analysis

I have not written here for a long time, and I just realized that I never published this exercise. It was from a class in functional analysis. Its purpose was to have us explicitly demonstrate the existence of a normed vector space that is not complete, i.e., not a Banach space, and of an invertible, bounded, linear operator whose inverse is not bounded.

Equip $\mathbb N$ with the discrete topology, let $c_c:=(C_c(\mathbb N), \|\cdot\|_\sup)$, i.e., the normed, linear space over $\mathbb C$ of continuous, compactly supported functions $\mathbb N\to\mathbb C$, and note that, since $f\in c_c$ is eventually zero,$$\|f\|_\sup=\max_{l\in\mathbb N}\{|f(l)|\}.$$For every $k\in\mathbb N$, define$$\begin{align}f_k:\mathbb N&\to\mathbb C\quad\text{by}\\l&\mapsto\begin{cases}l^{-1}&\text{ if }l\leq k\text{ and}\\0&\text{ otherwise},\end{cases}\end{align}$$note that $f_k\in c_c$, let $\varepsilon>0$, let $N>\varepsilon^{-1}$, let $m,n\in\mathbb N$, and assume, without loss of generality, that $m>n>N$. Then$$\|f_m-f_n\|_\sup=m^{-1}<N^{-1}<\varepsilon,$$i.e., $(f_k)_{k=1}^\infty$ is Cauchy.

Define $f:\mathbb N\to\mathbb C$ by $l\mapsto l^{-1}$, note that $f\not\in c_c$, let $\varepsilon>0$, let $N>\varepsilon^{-1}-1$, and let $k\in\mathbb N$ such that $k>N$. Then$$\|f_k-f\|_\sup=(k+1)^{-1}<(N+1)^{-1}<\varepsilon,$$i.e., $(f_k)_{k=1}^\infty$ converges to $f$.

Therefore, $c_c$ is not complete.

Define $$\begin{align}T:c_c&\to c_c\quad\text{by}\\f=(f_1,f_2,f_3,\dots)&\mapsto(\frac{f_1}{1},\frac{f_2}{2},\frac{f_3}{3},\dots).\end{align}$$

Let $f,g\in c_c$ and let $\lambda\in\mathbb C$. Then$$\begin{align}T(\lambda f+g)&=(\frac{\lambda f_1+g_1}{1},\frac{\lambda f_2+g_2}{2},\frac{\lambda f_3+g_3}{3},\dots)\\&=(\lambda\frac{f_1}{1}+\frac{g_1}{1},\lambda\frac{f_2}{2}+\frac{g_2}{2},\lambda\frac{f_3}{3}+\frac{g_3}{3},\dots)\\&=\lambda(\frac{f_1}{1},\frac{f_2}{2},\frac{f_3}{3},\dots)+(\frac{g_1}{1},\frac{g_2}{2},\frac{g_3}{3},\dots)\\&=\vphantom{\frac11}\lambda T(f)+T(g).\end{align}$$

Let $f=(f_1,f_2,\dots)\in c_c$ and note that$$\|T(f)\|_\sup=\max_{l\in\mathbb N}\{\frac{|f_l|}{l}\}\leq\max_{l\in\mathbb N}\{\frac{\|f\|_\sup}{l}\}=\|f\|_\sup.$$

Since $T$ is linear, $T$ is injective.

Let $f=(f_1,f_2,f_3,\dots)\in c_c$ and $g=(1f_1,2f_2,3f_3,\dots)\in c_c$. Then $T(g)=f$.

Define $S:c_c\to c_c$ by $f=(f_1,f_2,f_3,\dots)\mapsto(1f_1,2f_2,3f_3,\dots)$. Then $S\circ T(f)=T\circ S(f)=f$ so that $S=T^{-1}$.

Suppose that $M$ is such that $\|T^{-1}(f)\|_\sup\leq M\|f\|_\sup$ for every $f\in c_c$, let $k\in\mathbb N$ such that $k>M$, and define $g:\mathbb N\to\mathbb C$ by $l\mapsto1$ if $l=k$ and $l\mapsto0$ otherwise. Then$$\|T^{-1}(g)\|_\sup=\max_{l\in\mathbb N}\{|lg_l|\}=k=k\|g\|_\sup>M\|g\|_\sup,$$which is a contradiction.
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