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$\mathbb{CP}^1$ is the set of all one dimensional subspaces of $\mathbb{C}^2$. Let $(z,w)\in \mathbb{C}^2$ be non zero; its span is a point in $\mathbb{CP}^1$. Let $U_0=\{[z:w]:z\neq 0\}$ and $U_1=\{[z:w]:w\neq 0\}$, $(z,w)\in \mathbb{C}^2$,and $[z:w]=[\lambda z:\lambda w],\lambda\in\mathbb{C}^{*}$ is a point in $\mathbb{CP}^1$, the map is $\phi_0:U_0\rightarrow\mathbb{C}$ defined by $$\phi_0([z:w])=w/z$$ the map $\phi:U_1\rightarrow\mathbb{C}$ defined by $$\phi_1([z:w])=z/w$$ could any one tell me why the inverses of these maps are continuous?

$\bullet$ What is the Homeomorphism between $\mathbb{C}P^1$ and $\mathbb{C}\cup\{\infty\}$?

Fly by Night
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Myshkin
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  • First of all you need to tell us the topologies you are using on $\mathbb{CP}^1$ and $\mathbb{C}$. Your maps $\phi_i$ are continuous if the pre-image of each open set in $\mathbb{C}$ is open in $\mathbb{CP}^1.$ We need to know the topologies to know which sets are open and which are not. – Fly by Night Sep 03 '12 at 16:24
  • Could you check rick miranda's Algebraic Curves and riemann surface book? there is nothing specified. :( – Myshkin Sep 04 '12 at 03:13
  • I don't have that book I'm afraid. I've had a look elsewhere: $\mathbb{CP}^1$ is the set of orbits of the group action $\alpha : (\mathbb{C}^2)^* \times \mathbb{C}^* \to (\mathbb{C}^2)^$ where $\alpha(z,\lambda) := \lambda z.$ Thus, $\mathbb{CP}^1$ is the quotient space $(\mathbb{C^2})^ \backslash \mathbb{C}^$ , and so $\mathbb{CP}^1$ is given the quotient topology. Let $q : (\mathbb{C}^2)^ \to \mathbb{CP}^1$ be the quotient map where $q(z_1,z_2) := [z_1:z_2].$ A set $U \subseteq \mathbb{CP}^1$ is defined as being open if the pre-image $q^{-1}(U)$ is open in $(\mathbb{C}^2)^*$. – Fly by Night Sep 04 '12 at 16:09
  • Notice that this definition is still ambiguous because you need to define a topology on $(\mathbb{C}^2)^$ to be able to define a quotient topology on $\mathbb{CP}^1$. Although, it would most probably be the usual metric topology defined by the metric $d : (\mathbb{C}^2)^ \to \mathbb{C}$ where $d(z_1,z_2) := |z_1 - z_2|.$ – Fly by Night Sep 04 '12 at 16:13

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