Consider the usual spaces $SU(2)$ and $SO(3).$ I am trying to (formally) show that $\pi_1(SO(3))\simeq \mathbb Z_2.$
I was able to prove that $\pi_1(SO(3))$ has exactly two elements which I denote by $[c]$ and $[f]$. Here, the loop $c$ is the constant loop and $f$ is just some loop whose lift is a path connecting antipodal quaternions in $SU(2).$
To prove there is an actual isomorphism between these groups, I defined the application $\phi\colon \mathbb Z_2 \to \pi_1(SO(3))$ defined by $\phi(0) = [c]$ and $\phi(1) = [f].$
I was able to prove bijectivity quite easily (in fact, there isn't much to prove). My major question comes when I tried to prove that $\phi$ is a group homomorphism.
My attempt. To prove $\phi$ is an homomorphism, I considered all possible cases:
- $\phi(0 + 0) = \phi(0) = [c] = [c \cdot c] = [c][c] = \phi(0)\phi(0) - $ this follows directly from the fact that the fact that the homotopy class of the constant loop is the neutral element for the operation in the fundamental group.
- $\phi(0+1)=\phi(1)=[f]=[f\cdot c] = [f][c]=\phi(1)\phi(0) - $ for the same reason as point $1.$
- $\phi(1+1) = \phi(0) = [c]$ ?$=$?
My trouble is in point $3.$ I tried to evaluate $\phi(1)\phi(1)=[f \cdot f]$ but I cant show that $f \cdot f$ is homotopic to the constant loop, $c$. From results I have prior to this, it would suffice to show that the unique lift of $f \cdot f$ that start at $q$ (where $q$ is the positive antipodal quaternion) is a loop - but I also don't know how to show this.
Any help is apreciatted.
Nothing remains to show. You know that $\pi_1(SO(3))$ is a group with two element $e = [c]$ and $g = [f]$. Since $c$ is the constant loop, $e$ is the neutral element of $\pi_1(SO(3))$. The element $g$ has an inverse $g^{-1} \in \pi_1(SO(3))$, and since $g \ne e$, we must have $g^{-1} \ne e$, and thus $g^{-1} = g$. This automatically shows that your map $\phi$ is an isomorphism.