Note that we can embed $\Bbb RP^2$ into a topological space that is homotopy equivalent to $S^2$. This is because $\Bbb RP^2$ embeds in $\Bbb R^n$ for some large $n$ (which is true for every smooth manifold), and so $\Bbb RP^2$ embeds in $S^2\times \Bbb R^n$ which is homotopy equivalent to $S^2$.
My question is, can we embed $\Bbb RP^2$ as a subcomplex of a 2-dimensional cell complex that is homotopy equivalent to $S^2$? This means that regarding $\Bbb RP^2$ as a (2-dimensional) cell complex, we can attach more 0,1,2-cells to $\Bbb RP^2$ to get a space homotopy equivalent to $S^2$. I can't see whether this is true or not. Any hints?
Yes, there exists such a complex.
Hint: Attach a $2$-cell to $\mathbb{R}P^2$ via a homeomorphism $\mathbb{R}P^1 \cong S^1 = \partial D^2$. Equivalently, this can also be described as the mapping cone of the inclusion $\mathbb{R}P^1 \subset \mathbb{R}P^2$.
More generally, the mapping cone of the inclusion $\mathbb{R}P^{n-1} \subset \mathbb{R}P^n$ is a complex of dimension $n$ which is homotopy equivalent to $S^n$ and contains $\mathbb{R}P^n$ as a subcomplex.