Topological space with binary operation has abelian fundamental group

Question

I have been given this problem:

Let $\textit{X}$ be a topological space and $\mu : \textit{X} \times \textit{X} \rightarrow \textit{X}$ a binary operation. Show that if $\mu$ is continuous and $\textit{X}$ has an identity element $\textit{e}$ for $\mu$ then $\pi_1(\textit{X}, \textit{e})$ is abelian.

We want to show that the concatenation $\beta \star \alpha$ is path homotopic to $\alpha \star \beta$ $\forall \alpha, \beta \in \pi_1(\textit{X}, \textit{e})$ but I don't know how to do.

Thank you!

Answer 1

Let $\alpha,\beta:[0,1]\to X$ be two loops such that $\alpha(0)=\beta(0)=\beta(1)=\alpha(1)=e$. It suffices to show that $\alpha\star\beta=\beta\star \alpha$. Consider the space $[0,2]\times [0,2]$, and define the unique map $f:[0,2]\times [0,2]\to X$ such that $$f(x,y)=\begin{cases} \alpha(x) & \text{if }x\in [0,1]\text{ and }y=0 \\ \beta(y) & \text{if }x=0\text{ and }y\in [0,1]\\ \beta(x-1) & \text{if }x\in[1,2]\text{ and }y=0\\ \alpha(y-1) & \text{if }x=0\text{ and }y\in[1,2]\\ \mu(f(x,0),f(y,0)) & \text{always} \end{cases}$$ Observe that $f(-,0):[0,2]\to X$ and $f(0,-):[0,2]\to X$ are continuous, so the map $\chi: [0,2]\times [0,2]\to X\times X$ who takes $(x,y)\mapsto (f(x,0),f(0,y))$ is also continuous. It follows from the fact that $f=\mu\circ \chi$ that $f$ is itself continuous.

Now note that there exists a continuous map $\Delta:[0,1]\times [0,1]\to [0,2]\times [0,2]$ such that $\Delta:(x,0)\mapsto (2x,0)$ and $\Delta:(x,1)\mapsto (0,2x)$—I leave it to you to construct such $\Delta$.

The composition $[0,1]\times [0,1]\overset{\Delta}{\longrightarrow} [0,2]\times [0,2]\overset{f}{\longrightarrow}X$ gives the desired homotopy $\blacksquare$

Answer 2

Consider the group homomorphism induced by $\mu$ on the fundamental groups: $$f\colon \pi_1(X,e)\times\pi_1(X,e)\cong\pi_1(X\times X, (e,e))\xrightarrow{\mu_\star}\pi_1(X,e)$$

Furthermore, we have $f(1,\alpha)= \alpha = f(\alpha,1)$ for every $\alpha\in\pi_1(X,e)$, because for every representative $\alpha:S^1\to X$ the composition $$S^1\xrightarrow{(c_e,\alpha)}X\times X\xrightarrow{\mu}X$$ (where $c_e$ is the constant loop at $e$ representing the identity of $\pi_1(X,e)$) is equal to $\alpha$ (because $e$ is the unit of $\mu$).

Now, it is a general fact about groups that if a group $G$ admits a group homomorphism $f\colon G\times G\to G$ with $f(1,a)=a=f(a,1)$ for all $a\in G$ then $G$ is abelian. Because: $ab=f(a,1)f(1,b)=f(a1,1b)= f(1a,b1)= f(1,b)f(a,1)=ba$, where the second and fourth identities hold because $f$ is a group homomorphism.