Prove that for any point $p ∈ \Bbb RP_n$ , $\Bbb RP_n − p \simeq S^{n−1}$ Hope this isn't a duplicate.
I was trying the following Problem:

Prove that for any point $p ∈ \Bbb RP_n$ , $\Bbb RP_n − p \simeq S^{n−1} \text{ (homotopy equivalent)}$

But I am not quite sure whether the statement is true.
Like, for $n=1$,$\Bbb RP^1 \cong S^1$ (homeomorphic). Thus,$\Bbb RP^1 - p \cong S^1 - p \cong (0,1)$. On the other hand, $ S^0 = \{1,-1\}$ .
Now, since, $(0,1) \simeq *$, we have that $\pi_{0}(\Bbb RP^1 - p) =0$ i.e. trivial ! But, $\pi_{0}(S^0) = \Bbb Z/2\Bbb Z$ .
I am totally new to homotopy and fundamental groups, so please correct me if there is any mistake.
Anyway, Is the statement true for $\forall n \ge 2?$. In that case how to prove the statement?
Thanks in advance for help!
 A: $$
\newcommand{\rpn} {\Bbb R P^n}
$$
The statement you've got is ... almost true (at least for $n \ge 2$). And the case $n = 2$ is particularly misleading, because the true statement is that 
for any point $Q \in \rpn$, 
$$
\rpn - \{Q\} \simeq \Bbb RP^{n-1},
$$
but for $n = 2$, we have (as you observed) that this is the same as $S^{n-1}$. The problem is that if my version of the statement is true, yours cannot be, because in general $\rpn$ and $S^n$ are not homotopy equivalent. (For instance, their fundamental groups differ, for $n > 1$.)
To prove this (i.e. "my version") in general (for $n \ge 2$, because, as you observe, it's not true for $n = 1$), it's helpful to think of the double cover $p:S^n \to \rpn$. If you delete a point $P$ from $\rpn$, you might as well take the two preimages $p^{-1}(P)$ as the north and south poles of a coordinate system on $S^n$. Once you delete those from $S^n$, what you're left with is something that looks like a cylinder...which can retract onto its generating circle, which corresponds to the equator $S^{n-1} \subset S^n$. I like to think of each point on a globe "falling" towards the equator along its own line of longitude. (I supposed that if I were from the Southern Hemisphere, I'd think of them as "rising" towards the equator, or I'd turn my globe upside down...) The points whose "longitude line" isn't well defined are exactly the poles...which we don't have to consider! Yay!
And that's the insight you need as a way to get started. If you project that "retraction" back down into $\rpn$, you see each longitudinal retraction as sliding $\rpn - \{P\}$ onto the image (under $p$) of the equatorial $S^{n-1}$, 
which is the "equatorial" $\Bbb RP^{n-1}$. 
Now let's get explicit. I'm going to choose a coordinate system on $S^n$ in which the preimage of $P$ consists of the north and south poles, so that the homogeneous coordinates of $P$ are $[r, 0, \ldots, 0]$ for any nonzero $r$. 
Then I'm going to write points in $\rpn$ as unit vectors $(x_0, \ldots, x_n)$, where there's a sign ambiguity here that'll turn out not to matter. To make the sign ambiguity clear, I'll follow a kind of old-style convention and use colons: 
$$
Q = (x_0:\ldots:x_n)
$$
Note that for any such point $Q$, we can define $\theta(Q) = |\sin^{-1}(x_0)|$, which is independent of sign, hence well-defined. Roughly, it's the latitude of the preimage of $Q$ in $S^n$, but ignoring the north-vs-south latitude distinction. By negating all entries of $Q$ (hence leaving it unchanged) if necessary, we can assume that $x_0 \ge 0$, and can then write 
$$
Q = \sin(\theta(Q)) \cdot (1:0:\ldots:0) + (0:x_1: \ldots:x_n)
$$
and then, since the right hand term is a vector of length no greater than $\cos(\theta(Q))$, we can write
$$
Q = \sin(\theta(Q)) \cdot (1:0:\ldots:0) + \cos(\theta(Q))(0:u_1: \ldots : u_n)
$$
where $u_i = x_i / \cos(\theta(Q))$, and the vector $(0:u_1:\ldots:u_n)$ is a unit vector. 
I'll write all points in $\rpn$ in the form above, which we can condense by writing 
$$
Q = (\theta; \mathbf u) 
$$
as shorthand for the formula above. 
Now I'm going to define a deformation retraction $H: \rpn \times I \to \rpn$ as follows:
$$
H: \rpn \times I \to \rpn: ((\theta; \mathbf u), t) \mapsto ((1-t) \theta; \mathbf u).
$$
You can see that for $t = 0$, $H$ is just the identity. But for $t = 1$, the image of $H$ consists of points in $\Bbb RP^{n-1} \subset \rpn$. And for any point $Q' = (0, \mathbf u)$ that's already in $\Bbb RP^{n-1}$, we have
$$
H(Q', t) = Q'
$$
for all $t$. This shows that $\Bbb RP^{n-1}$ is a deformation retract of $\rpn - \{Q\}$, as required. 
A: Your argument for $n=1$ is almost correct, but you probably meant $\pi_0$ instead of $\pi_1$. I don't think the statement for $n \geq 2$ is true, but maybe I made a mistake. 
See $RP^n$ as $S^n$ with antipodal points identified. Deleting a point on $RP^n$ is the same as deleting the north/south poles on $S^n$ and then taking the quotient. It's easy to find a retraction of $S^n - \{N,S\}$ onto the equator $S^{n-1}$ which is invariant under the involution. This tells you that $RP^n - \{pt\}$ is homotopy equivalent to $S^{n-1}/\sim$ i.e $RP^{n-1}$.
