Show that the map $f:\mathbb{C}P^1 \rightarrow \mathbb{C}\cup \{\infty\}$ is a homeomorphism.

Question

Show that $$f:\mathbb{C}P^1 \rightarrow C\cup \{\infty\}$$ $$ [a:b]\mapsto b/a, \text{ if } a\neq0, \text{ else } \infty.$$ is a homeomorphism. Since $\mathbb{C}P^1$ is compact (is this trivial?) it suffices to show that $f$ is continuous and bijective. One can always choose a representative of the form $[1:b]$ for all elements of $\mathbb{C}P^1$ if $a\neq 0$. If $a=0$ we have $[0:b]=[0:1]$ which corresponds to $\infty$. So, that fact should give the bijective property. How can I show that $f$ is continuous? Is it enough to say $\lim_{a\to 0}|b/a|=0$ for a fixed $b$? Thanks for your help!

Answer 1

Notice that $\mathbb{C}P^1\cong\mathbb{S}^{3}/\mathbb{S}^1$, where $\mathbb{S}^3$ is the unit sphere of $\mathbb{R}^4$ endowed with the standard euclidean topology. Therefore, since $\mathbb{S}^3$ is compact, $\mathbb{C}P^1$ is compact as the direct image of the canonical projection, which is continuous by definition of the quotient topology. If you ask compact sets to be Hausdorff, then notice that $\mathbb{S}^1$ acts properly discontinuously on $\mathbb{S}^3$.

In order to show that $f$ is continuous, let $\pi\colon\mathbb{C}^2\setminus\{(0,0)\}\twoheadrightarrow\mathbb{C}P^1$ be the canonical surjection, then one has: $$(f\circ\pi)(a,b)=\frac{b}{a}$$ which is continuous between $\mathbb{C}^2$ and $\mathbb{C}\cup\{\infty\}$. Therefore, $f$ is continuous, indeed if $U$ is an open subset of $\mathbb{C}\cup\{\infty\}$, then $(f\circ\pi)^{-1}(U)=\pi^{-1}(f^{-1}(U))$ is open in $\mathbb{C}^2\setminus\{(0,0)\}$. Finally, by definition of the quotient topology, $f^{-1}(U)$ is open in $\mathbb{C}P^1$.