How do I prove that this is a Homeomorphism?

Solution 1:

The purpose of the statement in your question is to prove that $\mathbb{RP}^n$ is a topological manifold. To do so, it does not make sense "to use the structure of topological manifolds". You cannot use something what you have to show.

It is clear that the $U_k$ form an open cover of $\mathbb{RP}^n$. You know that $y_k$ is bijective, thus in fact you need to show that $y_k,y_k^{-1}$ are continuous.

Let $p : \mathbb R^{n+1} \setminus \{0\} \to \mathbb{RP}^n$ denote the quotient map. Then $V_k = p^{-1}(U_k) = \{ (x_0,\ldots,x_n) \mid x_k \ne 0 \}$ which is open in $\mathbb R^{n+1} \setminus \{0\}$. The restriction $p_k : V_k \to U_k$ of $p$ is a quotient map: Let $W \subset U_k$ be a set such that $p_k^{-1}(W)$ is open in $V_k$. We have $p_k^{-1}(W) = p^{-1}(W)$, thus this set is open in $\mathbb R^{n+1} \setminus \{0\}$. Hence $W$ is open in $\mathbb R^{n+1}$ and therefore open in $U_k$.

1. $y_k$ is continuous.
   The function $z_k : V_k \to \mathbb R^n, z_k(x_0,\ldots,x_n) = \dfrac{1}{x^k}(x^0,...,x^{k-1},x^{k+1},...,x_n)$, is continuous. Since $z_k = y_kp_k$, we conclude that $y_k$ is continuous.

2. $y_k^{-1}$ is continuous.
   The map $j_k : \mathbb R^n \to V_k, j_k(u_0,\ldots,u_{n-1}) = (u_0,\ldots,u_{k-1},1,u_k,\ldots,u_{n-1})$, is continuous. Thus $j'_k = p_kj_k : \mathbb R^n \to U_k$ is continuous. We have $y_k j'_k = id$, hence $j'_k = y_k^{-1}$.