Is it true that the product of not contractible loops are not contractible

Question

Is it true that the of product not contractible loops are not contractible, I think this can be proved by calculating the degree of the map, but I did not figure out how

Answer 1

It seems that you know about the fundamental group. So you might rephrase your question as: "if $x,y\in \pi_1(X)$ are non-trivial, is it true that $xy$ is non-trivial?".

But obviously, that is false is any group, since every element has an inverse. Precisely when $x$ and $y$ are inverse in $\pi_1(X)$, $xy$ is the trivial element, ie the class of a contractible loop.

So given a loop $\gamma$ in a topological space $X$, there is always a loop $\gamma'$ such that $\gamma\circ \gamma'$ is contractible, and actually the homotopy class of $\gamma'$ is uniquely determined by this fact.

For instance, in a circle, if $\gamma$ has index $n$, you would need to take $\gamma'$ with index $-n$ (visually, $\gamma'$ should "unwind" $\gamma$).