Some Problems with Fundamental Groups

Munkres Topology

53.3 Let $x_0$ and $x_1$ be points of a path-connected space $X$. Show that ${\pi }_1(X,x_0)$ is abelian if and only if for every pair $\alpha$ and $\beta$ of paths from $x_0$ to $x_1$, we have $\hat{\alpha }=\hat{\beta }$.

Proof) Assume ${\pi }_1(X,x_0)$ is abelian. For any loop $C$ based at $x_0$ and paths $\alpha$ and $\beta$ from $x_0$ to $x_1$, it suffices to show that $\overline{\alpha }\ast C\ast \alpha {\simeq }_p\overline{\beta }\ast C\ast \beta$. This boils down to simple computation:

$$\beta \ast \overline{\alpha }\ast C\ast \alpha \ast \overline{\beta }{\simeq }_p(\beta \ast \overline{\alpha })\ast C\ast (\alpha \ast \overline{\beta })$$

Note that $C$, $\beta \ast \overline{\alpha }$, and $\alpha \ast \overline{\beta }$ are loops based on $x_0$, and thus commute as ${\pi }_1(X,x_0)$ is abelian:

$$(\beta \ast \overline{\alpha })\ast C\ast (\alpha \ast \overline{\beta }){\simeq }_{p}C\ast (\beta \ast \overline{\alpha })\ast (\alpha \ast \overline{\beta }){\simeq }_{p}C\ast e_{x_0}{\simeq }_{p}C$$

Thus,

$$\overline{\beta }\ast C\ast \beta {\cong }_p\overline{\beta }\ast (\beta \ast \overline{\alpha })\ast C\ast (\alpha \ast \overline{\beta })\ast \beta {\simeq }_p\overline{\alpha }\ast C\ast \alpha$$

as desired. We now show the converse. Take any arbitrary two loops $C_1$ and $C_2$ based at $x_0$ and a path $\alpha$ from $x_0$ to $x_1$. Let $\beta$ be the path $C_2\ast \alpha$ from $x_0$ to $x_1$; then, by hypothesis, $\hat{\alpha }=\hat{\beta }$. Applying both functions to $C_1$ gives:

$$\overline{\alpha }\ast C_1\ast \alpha {\simeq }_p\overline{\alpha }\ast \overline{C_2}\ast C_1\ast C_2\ast \alpha \Rightarrow C_1{\simeq }_p\overline{C_2}\ast C_1\ast C_2\Rightarrow C_2\ast C_1{\simeq }_p{C}_1\ast C_2$$

This implies that $[C_2]\ast [C_1]=[C_1]\ast [C_2]$; as this holds for arbitrary loops, ${\pi }_1(X,x_0)$ is abelian, as desired.

53.6 Let $X$ be path-connected and $h$ be a continuous function from $X\to Y$ ($Y$ is an arbitrary space). Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$, and let $\beta =h\circ \alpha$. Show that

$$\hat{\beta }\circ (h_{x_0}{)}_{\ast }=(h_{x_1}{)}_{\ast }\circ \hat{\alpha }$$

That is, show that the fundamental group homomorphism induced by a continuous map is independent of base point up to group isomorphism.

Proof) For convenience, let $h(x_0)=y_0$ and $h(x_1)=y_1$. Note that for any loop $C$ based at $x_0$,

$$(\hat{\beta }\circ (h_{x_0}{)}_{\ast })(C)=\hat{\beta }([h(C)])=[\overline{\beta }\ast h(C)\ast \beta {]}_Y=(h_{x_1}{)}_{\ast }([\overline{\alpha }\ast C\ast \alpha ])=((h_{x_1}{)}_{\ast }\circ \hat{\alpha })(C)$$

53.7 Let $G$ be a topological group with operation $\cdot$ and identity $x_0$. Let $\Omega (G,x_0)$ denote the set of all loops in $G$ based in $x_0$. If $f,g\in \Omega (G,x_0)$, define $f\otimes g$ as $(f\otimes g)(s)=f(s)\cdot g(s)$. Show that $\otimes$ induces the canonical operator $\ast$ in the fundamental group ${\pi }_1(G,x_0)$, and that ${\pi }_1(G,x_0)$ is abelian.

Proof) We can clearly see $(\Omega (G,x_0),\otimes )$ is a group. On path-homotopy equivalence classes, $\otimes$ is well defined as if $F:I^2\to G$ is a path homotopy from $f$ to $f^{\prime }$ and $W$ is a path homotopy from $g$ to $g^{\prime }$, the composite map $H:I^2\to (F,W)\to F\cdot W$ is a path homotopy from $f\otimes g$ to $f^{\prime }\otimes g^{\prime }$. It, in fact, creates a group structure on ${\pi }_1(G,x_0)$: associativity transfers over from $\Omega (G,x_0)$, $e_{x_0}$ remains the identity, and the inverse of $C$ is the image of $C$ under the group inversion homeomorphism.

We now show that $\otimes$ and $\ast$ are the same on ${\pi }_1(G,x_0)$. This is pretty straightforward computation; for any two loops $C_1$ and $C_2$ based on $x_0$:

$$[C_1]\otimes [C_2]=[C_1\ast e_{x_0}]\otimes [e_{x_0}\ast C_2]=[C_1\ast C_2]=[C_1]\ast [C_2]$$

the same computation shows that the group must be abelian:

$$[C_1]\ast [C_2]=[C_1\ast e_{x_0}]\otimes [e_{x_0}\ast C_2]=[e_{x_0}\ast C_1]\otimes [C_2\ast e_{x_0}]=[C_2]\ast [C_1]$$