1  The Fundamental Group

1.1  (§51) Homotopy of paths

• $\mathrm{★}$ Def. $f,g:X\to Y$ are homotopic, $f\simeq g$, if there is a map $F:X\times I\to Y$ with $F(x,0)=f(x)$, $F(x,1)=g(x)$. $F$ is a homotopy.

• Notation. $f_t(x)=F(x,t)$; the rule $t\mapsto f_t$ is a path $I\to 𝒞(X,Y)$ from $f$ to $g$.

• Def. A path in $Y$ from $y_0$ to $y_1$: a map $f:I\to Y$ with $f(0)=y_0$, $f(1)=y_1$.

• $\mathrm{★}$ Def. Paths $f,g$ from $y_0$ to $y_1$ are path homotopic, $f{\simeq }_{p}g$, if there is $F:I\times I\to Y$ with $F(s,0)=f(s),F(s,1)=g(s),F(0,t)=y_0,F(1,t)=y_1.$ The endpoints stay fixed throughout.

• Lem. $\simeq$ and ${\simeq }_p$ are equivalence relations.

  Proof. Reflexive: constant homotopy. Symmetric: $\overline{F}(x,t)=F(x,1-t)$. Transitive: concatenate, $(F\star G)(x,t)=F(x,2t)$ for $t\le \frac{1}{2}$, $G(x,2t-1)$ for $t\ge \frac{1}{2}$; continuous by the pasting lemma since $F(x,1)=g(x)=G(x,0)$.

• Notation. $[f]$ = homotopy class; for paths, $[f]$ = path homotopy class.

• $\mathrm{★}$ Ex. Straight-line homotopy. For $f,g:X\to {ℝ}^2$ (or into any convex $Y\subset {ℝ}^n$):   $F(x,t)=(1-t)f(x)+tg(x)$. So any two maps into a convex set are homotopic; and if $f,g$ are paths with the same endpoints, this is a path homotopy.

• $\mathrm{★}$ Ex. In the punctured plane $Y={ℝ}^2-\{0\}$, with $f(s)=(\cos \pi s,\sin \pi s)$, $g(s)=(\cos \pi s,2\sin \pi s)$, $h(s)=(\cos \pi s,-\sin \pi s)$: $f{\simeq }_{p}g$ (the straight-line homotopy stays in $Y$), but the straight-line homotopy from $f$ to $h$ passes through the origin. In fact $f{\simeq ̸}_{p}h$ — proved later.

  Note. $f$ and $h$ are homotopic as maps if endpoints are released. That is why path homotopy, not homotopy, is the right relation here.

• $\mathrm{★}$ Def. For $f$ a path $x_0\to x_1$ and $g$ a path $x_1\to x_2$, the product

  Continuous by the pasting lemma.

• Lem. $f_0{\simeq }_p{f}_1$ and $g_0{\simeq }_p{g}_1$ $\Rightarrow f_0\ast g_0{\simeq }_p{f}_1\ast g_1$. Hence $[f]\ast [g]=[f\ast g]$ is well defined.

• $\mathrm{★}$ Thm. $\ast$ on path homotopy classes satisfies:

  1. Associativity: $([f]\ast [g])\ast [h]=[f]\ast ([g]\ast [h])$;

  2. Units: with $e_x(s)\equiv x$ the constant path, $[e_{x_0}]\ast [f]=[f]=[f]\ast [e_{x_1}]$;

  3. Inverses: with $\overline{f}(s)=f(1-s)$, $[f]\ast [\overline{f}]=[e_{x_0}]$ and $[\overline{f}]\ast [f]=[e_{x_1}]$.

  Proof. All three are reparametrisation homotopies. For (1) let the breakpoints slide: traverse $f$ on $0\le s\le (1+t)/4$, $g$ on $(1+t)/4\le s\le (2+t)/4$, $h$ thereafter. Draw the square with the two subdivision patterns and interpolate.

• Caution. $\ast$ is not a group operation on paths — only on path homotopy classes, and only when endpoints match.