Quotient space of $S^n$ and the projective plane

The quotient space on $S^n \times I$ obtained from equating $(x, 0) \sim (-x, 1)$ seems like it might have the same fundamental group as the projective plane, but I'm not entirely sure how to prove it. Is there a way to use van Kampen here?

The homeomorphism $x\mapsto -x$ of $S^n$ is homotopic to the identity of $S^n$ if and only if $n$ is odd. Thus, using my answer to you previous question, you know that whenever $n$ is odd, the mapping torus of the antipodal map of $S^n$ is homeomorphic to the mapping torus of the identity map of $S^n$, which is $S^n\times S^1$. Projective space $\Bbb{RP}^n$, for $n\geq 2$ has fundamental group $\Bbb Z/2\Bbb Z$, while for $n\geq 2$, $S^n\times S^1$ has fundamental group $\Bbb Z$. When $n=1$, then the fundamental group is isomorphic to $\Bbb Z\times\Bbb Z$. So at least when $n$ is odd, your claim cannot be correct.

EDIT. Suppose $n+1=2p$ is even, and identify the $\Bbb R^{2p}$ with $\Bbb C^p$. Then there is a homotopy through homeomorphisms linking the identity to the antipode given by $$H((z_1,\dots,z_p),t)=(e^{i\pi t}z_1,\dots,e^{i\pi t}z_p)$$

EDIT. The mapping cylinder $M_\phi$ associated to a homeomorphism of an orientable manifold $X$ is (a) a manifold, (b) orientable iff $\phi$ preserves the orientation of $X$. Since the antipode reverses the orientation whenever $n$ is odd (which we assume below), this shows that the mapping cylinders of the identity and of the antipode aren't homeomorphic, as the first one isn't orientable, while the second one is.

In any case, you can check this explicitely by moving a frame $\mathcal{B}_{u}\equiv$ basis of the tangent space to $M_{\text{antipode}}$ at $c(u)$ along the continuous loop $$c(u)=\begin{cases}0\leq t\leq \frac12:&[(N,2u)]\\ \frac12\leq t\leq 1:&c(u)\text{ links }[(-N,0)]\text{ to }[(N,0)]\end{cases}$$ (where $N=(0,\dots,0,1)$, and the second part of the loop could follow a great circle drawn on $S^n$). Since the antipode reverses orientation, the two frames $B_0,B_1$ define two different orientations of the tangent space at $[(N,0)]$, and $M_{\text{antipode}}$ is non-orientable.

• But as you mentioned before, is that homotopy a homotopy through homeomorphisms? I don't think the same homotopy will work because the great circles might intersect – Yon Kim Jan 19 '15 at 11:03
• Ok that makes sense. I actually think the answer is the same for $n$ even but not sure how to show it – Yon Kim Jan 19 '15 at 12:43
• Can this be done using van Kampen? Take $U$ to be half the ball and $V$ the other half. Since $U \cap V$ has one less dimension we know it's fundamental group – Yon Kim Jan 19 '15 at 15:11
• I doubt it: I don't thinkthe mapping cylinder is orientable when $n$ is even. – Olivier Bégassat Jan 19 '15 at 16:49
• Is there anyway to get this for $n$ even? I know that the quotient map on antipodes is homeomorphic to the projective plane but I'm not sure that helps – Yon Kim Jan 19 '15 at 16:59