$\mathbf{R}P^{n}/ \mathbf{R}P^{n-2} \simeq S^n \vee S^{n-1}$ iff $n$ is odd Prove that $\mathbf{R}P^{n}/ \mathbf{R}P^{n-2} \simeq S^n \vee S^{n-1}$ iff $n$ is odd. One side is solved just calculating the homology groups of both sides. I can't see if $n$ is odd how to show this. Any help is appreciated.
Please use only homology (try not to use cohomology, Hopf Theorem etc.) 
 A: When I see a statement about spheres depend on dimension, I immediately think about the degree of the antipodal mad. If $n$ is odd, then it is homotopic to the identity.
Now let's imagine $\mathbb{R}P^n / \mathbb{R}P^{n-2}$. We know that $\mathbb{R}P^n$ is a disk attached to $\mathbb{R}P^{n-1}$, so this quotient is a disk attached to $\mathbb{R}P^{n-1} / \mathbb{R}P^{n-2}$. This latter thing is just $S^{n-1}$. Now how should a wedge sum come about? Well it should come about from attaching a disk using the constant map on the boundary. This is equivalent (in the homotopy sense) to attaching a disk using a degree 0 map.
So we hope to prove that when $n$ is odd this map has degree 0. The specific map we get is the result of the projection $S^{n-1} \rightarrow \mathbb{R}P^{n-1}$ followed by quotienting out the equatorial copy of $\mathbb{R}P^{n-2}$. Well, $n-1$ is even, so $H_{n-1}(\mathbb{R}P^{n-1})=0$, meaning this composition is degree 0. So we have the homotopy equivalence in question.
The fact that this homology group is trivial is really a reflection of degrees. The projection map can be factored both through the identity (obviously) and the antipodal map, which means that if the degree of the antipodal map is -1, the projection induces the zero map on homology. Then you just mess around with cellular homology. 
