I'm dealing with the following question:
Consider the inclusion $i : \mathbb{R}P^{2} \rightarrow \mathbb{R}P^{3}$ induced by the inclusion of $S^{2}$ into $S^{3}$ as its equator. Show that $i$ is not homotopic to a constant map.
It's a practice problem, but I'm stuck on how to show this. I feel as if it's a very simple step that I'm missing?
It is well-known that $\mathbb RP^n$ has a CW-structure with one cell in each dimension. In fact, $\mathbb RP^n$ is obtained from $\mathbb RP^{n-1}$ by attaching an $n$-cell. See CW complex structure of the projective space $\mathbb{RP}^n$. Thus the $k$-skeleton of $\mathbb RP^n$ is $\mathbb RP^k$. By cellular approximation we see that the inclusion $i : \mathbb RP^2 \to \mathbb RP^3$ induces an isomorphism $$i_* : \pi_1(\mathbb RP^2) \to \pi_1(\mathbb RP^3).$$ This is just a special case of the fact the the inclusion $i : X^2 \to X$ of the $2$-skeleton $X^2$ of a CW-complex $X$ induces an isomorphism $i_* : \pi_1(X^2) \to \pi_1(X)$.
But is well-known that $\pi_1(\mathbb RP^2) \approx \mathbb Z_2$, thus $i$ cannot be null-homotopic (in that case we would have $i_* = 0$ which is not an isomorphism).
