Define a continuous map $\ell:(I,\partial I)\to (SO(3),1)$ by $\ell(t) = \left( \begin{array}{ccc} \cos 2\pi t & -\sin 2\pi t & 0 \\ \sin 2\pi t & \cos 2\pi t & 0 \\ 0 & 0 & 1 \end{array} \right) $ then concretely find homotopy that give $[\ell]^2=1\in\pi_1\text{(SO}(3),1)$
I have to solve this as an assignment however still have no idea how to do it because I could not catch up with the lecture(Honestly,I still don't know any properties of fundamental group and which should be applied here)
Could anyone show me how to do this step-by-step(also which properties,definitions are applied)? (Mayer-Vietoris,Fundamental group's properties,H-space,Van kampen..was taught recently so I wonder they can be used here) Really thanks for your help.
I would try and write this as a comment but I don't have that ability yet.
To find the answer you only need to use the definition of the fundamental group, I would say that it requires you to think more about the space $SO(3)$ than properties of $\pi_1$. But let me push you in the right direction.
The question is asking you to show that the class $[\ell]^2$ is the same as the class of the constant loop at $1$. The first question is, what is the class $[\ell]^2$? The multiplication in $\pi_1$ is given by $[f]*[g]= [g\circ f]$ where $$ (g\circ f) (t)=\begin{cases} f(2t) & 0 \leq t \leq \tfrac{1}{2} \\ g(2t-1) & \tfrac{1}{2} \leq t \leq 1 \end{cases}$$
So the class $[\ell]^2$ is the same as $[\ell\circ \ell]$, i.e. the class represented by the loop $\ell \circ \ell$. Recall that $SO(3)$ can be thought of as the group of rotations on $\mathbb{R}^3$, so can you see what this loop is geometrically? The question still remains to show $[\ell \circ \ell]$=$[1]$, this boils down to finding a homotopy between $\ell \circ \ell$ and $1$, i.e. a map $H:I\times I\rightarrow SO(3)$ with $(0,t)=1=(1,t)$ and $H(\theta,0)=\ell \circ \ell, H(\theta,1)=1$. This takes some time to find, but I would hint at looking at paths in $SO(3)$ that are similar to $\ell$. I hope this helps.