Show P^1 is homemorphic to S^1

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Show P^1 is homemorphic to S^1
I know I need to prove there is a function satisfying it's 1-1 ,onto,continuous,and the inverse of function is continuous.However, I can't find it.Please help!
 
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What is your definition of P^1?

Knowing about stereographic projection might be helpful as well, so look that up.
 
P^1 is the set of all line in R^2(or R^2\(0,0), I forget which one is right) through the origion.
 
jack8572357 said:
P^1 is the set of all line in R^2(or R^2\(0,0), I forget which one is right) through the origion.

Can you attach to a line in \mathbb{R}^2 a real number?? For example, given a line through the origin ax+by=0, I can look at the slope -b/a (works if a is nonzero).

So, that gives a function between P^1 except one point and \mathbb{R}. Then apply stereographic projection.
 
Would you say it in detail?I have no idea about it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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