The Fundamental Group

We prove each part by exhibiting explicit continuous homotopies; this is the standard technique for path-homotopy statements and it is what makes the whole theorem non-circular — every claim is verified by writing down a formula for $H(s,t)$ and checking the four defining conditions directly from Homotopy of paths, the fundamental group.

(1) Equivalence relation. Reflexive: $H(s,t) = f(s)$ (constant in $t$) is continuous (a composition of $f$ with the projection $[0,1]\times[0,1]\to[0,1]$) and satisfies $H(s,0)=H(s,1)=f(s)$, $H(0,t)=f(0)$, $H(1,t)=f(1)$ for all $t$. So $f \simeq_p f$. Symmetric: if $H$ is a path homotopy from $f$ to $g$, define $H'(s,t) = H(s,1-t)$. This is continuous as a composition of $H$ with the continuous map $(s,t)\mapsto(s,1-t)$. Then $H'(s,0)=H(s,1)=g(s)$, $H'(s,1)=H(s,0)=f(s)$, and $H'(0,t)=H(0,1-t)=f(0)=g(0)$ (endpoints of $f,g$ agree since $f\simeq_p g$ requires equal endpoints), similarly $H'(1,t)=f(1)=g(1)$. So $g \simeq_p f$. Transitive: suppose $H$ is a path homotopy $f\simeq_p g$ and $K$ is a path homotopy $g \simeq_p h$. Define

On the overlap $t=1/2$ both formulas give $H(s,1) = g(s) = K(s,0)$, so $J$ is well defined; it is continuous because $[0,1]\times[0,1]$ is the union of the two closed sets $[0,1]\times[0,1/2]$ and $[0,1]\times[1/2,1]$ on which $J$ agrees with continuous functions and the two definitions agree on the (closed) intersection — this is the pasting lemma for closed sets. Checking boundary values: $J(s,0)=H(s,0)=f(s)$, $J(s,1)=K(s,1)=h(s)$, $J(0,t)$ equals $H(0,2t)=f(0)$ or $K(0,2t-1)=g(0)=f(0)$ in either range, similarly $J(1,t)\equiv f(1)$. So $f \simeq_p h$.

(2) Well-definedness of $*$. Suppose $H$ is a path homotopy $f \simeq_p f'$ and $K$ is a path homotopy $g \simeq_p g'$ (all loops based at $x_0$, so all boundary values equal $x_0$). Define

At $s = 1/2$: $H(1,t) = f(1) = x_0 = g(0) = K(0,t)$ for every $t$, since $H, K$ are homotopies of based loops, so the two formulas agree on the overlap and $L$ is continuous by the pasting lemma. One checks $L(s,0) = (f\cdot g)(s)$, $L(s,1) = (f'\cdot g')(s)$, and $L(0,t) = H(0,t) = x_0$, $L(1,t) = K(1,t) = x_0$ for all $t$. Hence $f\cdot g \simeq_p f'\cdot g'$, so $*$ is well defined on homotopy classes.

(3) Associativity. The functions $(f\cdot g)\cdot h$ and $f \cdot (g\cdot h)$ are generally different functions $[0,1]\to X$ — they run $f, g, h$ at different speeds and on different subintervals — but they are path-homotopic via an explicit reparametrization. Concretely, $(f\cdot g)\cdot h$ traverses $f$ on $[0,1/4]$, $g$ on $[1/4,1/2]$, $h$ on $[1/2,1]$ (each at quadruple/double speed as appropriate), while $f\cdot(g\cdot h)$ traverses $f$ on $[0,1/2]$, $g$ on $[1/2,3/4]$, $h$ on $[3/4,1]$. Define the piecewise-linear, continuous, surjective reparametrizing map $\varphi_t : [0,1]\to[0,1]$ interpolating between these two break-point schedules as $t$ goes from $0$ to $1$: explicitly let

which is continuous in $(s,t)$ jointly (each piece is a jointly continuous rational function of $(s,t)$ on its domain with $t\in[0,1]$ keeping denominators bounded away from $0$, and the pieces agree at the two junctions by direct computation), with $\varphi_0(s)$ giving the break points $1/4, 1/2$ (matching $(f\cdot g)\cdot h$) and $\varphi_1(s)$ giving break points $1/2,3/4$ (matching $f\cdot(g\cdot h)$), and $\varphi_t(0)=0,\varphi_t(1)=1$ for every $t$. Setting

gives a continuous map (composition of continuous maps) with $H(s,0) = ((f\cdot g)\cdot h)(s)$, $H(s,1) = (f\cdot(g\cdot h))(s)$ (direct substitution and comparison of the piecewise formulas on matching subintervals), and $H(0,t) = ((f\cdot g)\cdot h)(0) = f(0) = x_0$, $H(1,t) = x_0$ for all $t$. This is a path homotopy, so $(f\cdot g)\cdot h \simeq_p f\cdot(g\cdot h)$, i.e. $([f]*[g])*[h] = [f]*([g]*[h])$. (In practice one usually just invokes the general reparametrization lemma: if $\varphi:[0,1]\to[0,1]$ is continuous with $\varphi(0)=0,\varphi(1)=1$, then $f\simeq_p f\circ\varphi$ via the straight-line homotopy $H(s,t) = f\big((1-t)s + t\varphi(s)\big)$, which is continuous since it is a composition of continuous maps, and one checks the boundary conditions directly. Associativity, the identity laws, and the coherence of concatenation with reparametrized loops all reduce to this lemma plus the pasting argument above; we gave the fully explicit $\varphi_t$ above to make the non-circularity of the argument completely transparent.)

(4) Identity. We show $[f]*[e_{x_0}] = [f]$; the other equality is symmetric. Both $f \cdot e_{x_0}$ and $f$ have the same image and are related by the reparametrization

since $(f\cdot e_{x_0})(s) = f(\varphi(s))$ for all $s$ (direct check on each subinterval, using $e_{x_0}(2s-1)=x_0=f(1)$). Here $\varphi$ is continuous (pasting lemma, the two pieces agree at $s=1/2$: both give $1$), with $\varphi(0)=0,\varphi(1)=1$. By the reparametrization lemma, $f \circ \varphi \simeq_p f$ via $H(s,t) = f((1-t)s + t\varphi(s))$; continuity of $H$ is immediate as a composition, and $H(0,t)=f(0)=x_0$, $H(1,t) = f(1)=x_0$ for all $t$ since $\varphi(0)=0,\varphi(1)=1$. Hence $f\cdot e_{x_0} \simeq_p f$, so $[f]*[e_{x_0}]=[f]$.

(5) Inverses. We must show $f \cdot \bar f \simeq_p e_{x_0}$ (the proof that $\bar f \cdot f \simeq_p e_{x_0}$ is identical with the roles swapped). Define

At $s=1/2$ both formulas give $f(1-t)$, so $H$ is well defined and continuous by the pasting lemma (each piece is a composition of $f$ with a jointly continuous function of $(s,t)$). We check the four boundary conditions:

• $H(s,0)$: for $s\le 1/2$, $H(s,0) = f(2s) = (f\cdot\bar f)(s)$ since $f\cdot\bar f$ on $[0,1/2]$ is $f(2s)$; for $s\ge 1/2$, $H(s,0) = f(2(1-s)) = f(2-2s)$, and $(f\cdot \bar f)(s)$ for $s \ge 1/2$ is $\bar f(2s-1) = f(1-(2s-1)) = f(2-2s)$ — matches.
• $H(s,1) = f(0) = x_0$ for all $s$ (both branches), which is $e_{x_0}(s)$.
• $H(0,t) = f(0) = x_0$ for all $t$.
• $H(1,t) = f(0) = x_0$ for all $t$ (second branch at $s=1$ gives $f(2\cdot 0\cdot(1-t)) = f(0)$).

So $H$ is a path homotopy from $f\cdot \bar f$ to $e_{x_0}$, giving $[f]*[\bar f] = [e_{x_0}]$.

Conclusion. By (2), $*$ is a well-defined binary operation on $\pi_1(X,x_0)$; by (3) it is associative; by (4) $[e_{x_0}]$ is a two-sided identity; by (5) every element has a two-sided inverse. These are exactly the group axioms, so $(\pi_1(X,x_0), *)$ is a group. $\blacksquare$

We use the reparametrization/pasting technique established above, applied to concatenations of $\gamma$, $\bar\gamma$, and loops. Well-defined on classes: if $f \simeq_p f'$ (both loops at $x_0$) via homotopy $H$, then $\bar\gamma \cdot f \cdot \gamma \simeq_p \bar\gamma \cdot f' \cdot \gamma$ via the concatenation-of-homotopies construction from part (2) of the theorem above (concatenate the constant homotopy on $\bar\gamma$, then $H$ reparametrized into the middle third, then the constant homotopy on $\gamma$); this is a direct three-piece pasting-lemma argument identical in spirit to the proof of part (2), so $\hat\gamma([f])$ depends only on $[f]$. Homomorphism: We must check $\hat\gamma([f]*[g]) = \hat\gamma([f]) * \hat\gamma([g])$, i.e.

The right side traverses $\bar\gamma, f, \gamma, \bar\gamma, g, \gamma$ each at appropriate multiplied speed, with the middle segment $\gamma\cdot\bar\gamma$ path-homotopic to the constant loop $e_{x_1}$ by part (5) of the theorem above (with $\gamma$ in place of $f$, noting $\overline{\bar\gamma} = \gamma$, so $\gamma \cdot \bar\gamma \simeq_p e_{x_1}$; here we use $\gamma\cdot\bar\gamma$ is a loop at $x_1$). Inserting this path-homotopy into the six-fold concatenation, using part (2) to replace a segment of a concatenation by a path-homotopic segment while fixing the rest, and then using the identity law (part 4) to delete the resulting constant loop, and reparametrizing (reparametrization lemma) the remaining $\bar\gamma, f, \gamma$ back to standard thirds, yields

Hence $\hat\gamma$ is a homomorphism. Bijectivity: Let $\hat\gamma^{-1}\!: \pi_1(X,x_1)\to\pi_1(X,x_0)$ be defined the same way using the reverse path $\bar\gamma$ in place of $\gamma$, i.e. $\hat\gamma^{-1}([h]) = [\gamma \cdot h \cdot \bar\gamma]$. Then for a loop $f$ at $x_0$,

Using $\gamma\cdot\bar\gamma \simeq_p e_{x_0}$ (part 5) at both ends and the identity law (part 4) to cancel them (via the same "replace a segment, then delete constant loop" argument as above), this reduces to $[f]$. So $\hat\gamma^{-1}\circ\hat\gamma = \mathrm{id}$, and symmetrically $\hat\gamma \circ \hat\gamma^{-1} = \mathrm{id}$. Hence $\hat\gamma$ is a bijective homomorphism, i.e. a group isomorphism. $\blacksquare$