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Let $i:\mathbb{R}P^1 \to \mathbb{R}P^n$ be given by $[x_0,x_1] \mapsto [x_0,x_1,0,\ldots,0]$, with $n \ge 2$. Consider the fibration $v:S^1 \to \mathbb{R}P^1$ given by $(x_0,x_) \mapsto [x_0,x_1]$ (so the fiber is $S^0$). Show that $[i \circ v]$ is a non-trivial element of $\pi_1(\mathbb{R}P^n,\ast)$. (The $[a,b]$ notation refers to the equivalence class)

I am unsure where to start here. Firstly, $\pi_1(\mathbb{R} P^n,\ast)$ is $\mathbb{Z}/2\mathbb{Z}$, and $\mathbb{R} P^1 \simeq S^1$ so that $v$ is really an endomrphism.

I have been looking at the cohomlogy ring strucuture on $H^*(\mathbb{R}P^n,\mathbb{Z}/2\mathbb{Z})$ so it seems likely we need to use this here. Previously I have shown that for $n>m \ge 1$, and the embedding $i:\mathbb{R}P^m \hookrightarrow \mathbb{R}P^n$ we have that $i^*:H^q(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z}) \to H^q(\mathbb{R}P^m;\mathbb{Z}/2\mathbb{Z})$ is an isomorphism when $q \le m$, given by $i^*(\alpha_n) = \alpha_m,$ where $\alpha_n$ is the non-zero element of $H^1(\mathbb{R}P^n,\mathbb{Z}/2\mathbb{Z})$. As stated above I don't really know where to start. More specifically, how can I connect the homotopy and cohomology aspects of the question? (By definition, the map $v$ has the homotopy lifting property with respect to every space - do I use this?)

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    @user8268 - oh yes - because the map $v$ has degree 2? (I had forgotten that)2011-05-24

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