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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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I'm a bit confused: if I understand you correctly, $i\circ v$ is equa to the composition $S^1\to S^n\to \mathbb{R}P^n$ ($S^1\to S^n$ is the inclusion and $S^n\to \mathbb{R}P^n$ the quotient by $\pm1$). Since $S^n$ is $1$-connected for $n>1$, we get the trivial element of $\pi_1(\mathbb{R}P^n)$. – user8268 May 24 '11 at 7:32
@user8268 - sorry, maybe it is not too clear! (probably because I was using $i$ for two different things). For this question $i:\mathbb{R}P^1 \to \mathbb{R}P^n$ is the standard embedding, and $v:S^1 \to \mathbb{R}P^1$ is the fibration. – Juan S May 24 '11 at 7:35
@user8268: But I guess your commment will still hold. Can you explain the last sentence? – Juan S May 24 '11 at 7:36
@Qwirk: $i:S^1=\mathbb{R}P^1\to\mathbb{R}P^n$ gives you the non-trivial element of $\pi_1(\mathbb{R}P^n)$, but $[i\circ v]=2[i]=0\in\pi_1(\mathbb{R}P^n)$. But I might be missing something. – user8268 May 24 '11 at 7:41
Unless there's some formal catch, I agree with @user8268. The circle $S^1$ is wrapped "twice" - from a point to the opposite point of the higher-dimensional sphere and back. This is a trivial path/submanifold both in homotopy and homology, isn't it? That's twice of the generator of the $Z_2$ group. – LuboŇ° Motl May 24 '11 at 7:43

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