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Let $\mathbb{P^1}$ be the set of all lines $\mathbb{R^2}$ passing through the origin. And by the definition of the unit circle $S^1 = \left[0,1\right]/\sim$ , $\left[0,1\right]$ are identified.

What does it mean that $\mathbb{P^1}$ identifies with $S^1$?

What I supposed is that by definition $ S^1 \subset \mathbb{P^1}$.

Am I missing the point? What does it mean by "identifies with"?

Phosphene
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  • Hint: If two points are "identified" they are to be treated as though they were the exact same point. A way of looking at this is that you take the two points and glue one on top of the other (take the unit square $[0,1]\times[0,1]$ and identify the left and right hand sides. You should be able to see (use a sheet of paper if it helps) that this produces a cylinder). – postmortes May 12 '21 at 07:29
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    $\mathbb{RP}^1$ identifies with $\mathbb{S}^1$ means that they are homeomorphic – ocsecnarf ittorettul May 12 '21 at 07:34
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    @postmortes I can see how two opposites sides of a unit square can form a cylinder when the opposite sides are glued together, but the notation is somewhat new and a bit confusing to me. How does $[0,1]\times[0,1]$ creat a unit square exactly? I have to consider $[0,1]\times[0,1]$ as two lines, the first line creates a circle, and the other one "slides through" that circle creating a cylinder correct? How is this sliding effect called, spanning? Sorry, I am a complete beginner as you can see. Or is it two parallel lines gluing together? – Phosphene May 12 '21 at 08:03
  • $[0,1]\times[0,1]$ is the set ${(x,y) \mid x\in[0,1] \mbox{ and } y\in[0,1] }$, so it's every point in that square. When we identify points they're the same, so if you go left from $(0,0)$ you move to $(\epsilon,0)$ for $\epsilon >0$ and if you go right you move to $(1-\epsilon ,0)$ – postmortes May 12 '21 at 08:24
  • Possibly of interest? – Andrew D. Hwang May 12 '21 at 13:53

3 Answers3

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Consider $C=\{(x,y):x^2+y^2=1, x\geq 0, \}$, it is the half-circle it is an interval. every line in $\mathbb{R}^2$ which is different of $\{y=0\}$ meets this half-circle in exactly one point. The line $\{y,0\}$ meets $C$ in $(-1,0)$ and $(1,0)$. This defines a bijection between the projective plane and the quotient of $C$ with $(1,0)$ and $(-1,0)$ identified which is the circle.

Tsemo Aristide
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You already know by definition, $\Bbb RP^1$ is homeomorphic to $S^1/x\sim -x$ where $x\in S^1$.

Hint: 1. We have a commutative diagram $$\require{AMScd} \begin{CD} D^1 @>i>> S^1\\ @Vf VV @VVgV\\ D^1/\sim @>k>> S^1/\sim \end{CD}$$

where $i$ is a natural inclusion map and $k$ is an induced map that makes the diagram commutes and $f,g$ are quotient maps. Here, $D^1/\sim$ is just an identification of two endpoints. Using the definition of quotient topology, show that $k$ is a continuous map.

  1. Recall the theorem: a continuous bijective map $f:X\to Y$ where $X$ is compact and $Y$ is Hausdorff is a homeomorphism.

  2. Conclude $D^1/\sim\ \cong\ S^1/\sim$.

Remark. Using the same argument, we can show that $D^n/\sim\ \cong\ S^n/\sim$

one potato two potato
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Take the line y=1 in $\mathbb{R^2}$. Then, clearly you can define a bijection between this line and the set of lines through the origin without x=0 (every such line cuts y=1). So, the whole set of lines through the origin is in bijection with $\mathbb{R}$ plus a point, which itself is one-to-one with the circle. Now you can also show that this is not only a bijection but also an homeomorphism if you define a certain natural topology on the set of lines through the origin.

huurd
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  • what is meant by " $\mathbb{R}$ plus a point"? Shouldn't it be minus a point? I am taking a first-year geometry course, so I am not yet acquainted with topology, unfortunately. – Phosphene May 12 '21 at 07:58
  • You have to make a drawing while reading my previous answer, you will understand much better. $\mathbb{R}$ plus a point because y=1 is in bijection with $\mathbb{R}$, but you have to add another point corresponding to x=0 to get the bijection with the set of lines through the origin. – huurd May 12 '21 at 08:27