Let $\mathbb{R}P^2$ denote the real projective plane and suppose that $a$ is a non-contractible loop in $\mathbb{R}P^2$. Since $\pi_1(\mathbb{R}P^2)=\mathbb{Z}_2$, $a^2$ is contractible. I am trying to figure out the following:
Let $i:\mathbb{R}P^2\to\mathbb{R}P^3$ be the inclusion map (e.g. we obtain $\mathbb{R}P^3$ by identifying antipodal points of $\mathbb{S}^3$, so we can obtain $\mathbb{R}P^2$ as quotient of the subspace of $\mathbb{S}^3$ where the $4th$ coordinate is constantly 0). If we now see $a$ as a loop in $\mathbb{R}P^3$, does it become contractible? or does it remain non-contractible? In other words, is $i_*[a]=0$, where $i_*$ is the induced group homomorphisms between the fundamental groups?
I have little understanding of the real projective space, I don't even know how to compute its fundamental group (although I'm trying for a Seifert-van Kampen application) and it is hard for me to conceptualize the loop $a$ in $\mathbb{R}P^3$. Any intuitive or explicit hints and answers to help me understand this are greatly appreciated.
It is not contractible to see this consider the fibration $p_n:S^n\rightarrow \mathbb{R}P^n$,
A contractible loop $c:\rightarrow \mathbb{R}P^n$, lifts to a contractible loop of $S^n$ since for a Serre fibration homotopies can be lifted.
The non contractible loop of $\mathbb{R}P^n$ can be constructed by taking a path $c$ in the upper hemisphere of $S^n$ wich meets the equator in two points which are identified by $p_n$, the embedding $\mathbb{R}P^n\rightarrow \mathbb{R}P^{n+1}$ result from an embedding $i_n:S^n\rightarrow S^{n+1}$ and $i_n(c)$ is a non closed segment such that $p_n(i_{n+1}(c))$ is a loop, so it is not contractible since its lift is not a loop.