$\mathbf {The \ Problem \ is}:$ Show that if n is odd, then $X =\frac {RP^n}{RP^{n-2}}$$\simeq S^n \vee S^{n-1} =Y.$
$\mathbf {My \ approach}:$ We know $RP^n =\frac {S^n}{C_2},$ then is it true that $ \frac {\frac {S^n}{C_2}}{\frac {S^{n-2}}{C_2}} \cong \frac {S^n}{S^{n-2}}$ ?
If it is true, then as $S^{n-2}$ is a contractible subcomplex of $S^n,$ then by this image(pg 14, Hatcher's book) ; we are done .
A small hint is warmly appreciated , thanks in advance .