How to construct a map from S^2 to RP^2 with covering time being unity?

[wdlang]

it is easy to construct a map from S^2 to S^2, with covering time being unity

but how to do the similar task on the projected manifold RP^2=S^2/Z_2?

i tried to use the stereographical trick

the points on the lower half semisphere are projected onto the plane

the problem is that the infinity point on the plane is mapped to the circle, which is not a single point on RP^2.

 

[quasar987]

It is not clear to me what you are trying to get. What is a "map with covering time being unity"?

 

[wdlang]

It is not clear to me what you are trying to get. What is a "map with covering time being unity"?

we know the homotopy group

\pi_2(RP^2)=Z

i want to construct an element with the covering time being 1

 

[quasar987]

Still don't get it. An element of \pi_2(RP^2) is the homotopy class of a map S²-->RP². What is this buisiness about "covering time"?

 

[wdlang]

Still don't get it. An element of \pi_2(RP^2) is the homotopy class of a map S²-->RP². What is this buisiness about "covering time"?

you can call it the winding number

 

[quasar987]

You mean the degree?!? http://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping

But then RP² is non orientable...so how are you defining the degree of map S²-->RP²?

Afaik, the degree is only defined mod 2 in this case.

 

[wdlang]

You mean the degree?!? http://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping

But then RP² is non orientable...so how are you defining the degree of map S²-->RP²?

Afaik, the degree is only defined mod 2 in this case.

i do not know this at all

but could we construct a mapping in the class of [1]?

if we use the identity map to map s2 to s2 and then project to rp2

i guess the mapping is in the class of [2].

 

[quasar987]

So now it seems like you're talking about finding a map f:S²-->RP² that is a generator of $$\pi_2(RP^2)=Z$$.

But that is easy given the usually way of determing $$\pi_2(RP^2)$$.

Namely, under the projection (or quotient) map pi:S²-->RP², S² is a fiber bundle over RP^2 of fiber S^0={-1,1}. So there is a long exact sequence of homotopy groups which around pi_2 looks like:

$$0=\pi_2(S^0)\rightarrow \pi_2(S^2)\stackrel{\pi_*}{\rightarrow} \pi_2(RP^2)\rightarrow \pi_1(S^0)=0$$

So $$\pi_*$$ is an isomorphism. And what does it do? It takes the class of the identity map [id:S^2\rightarrow S^2] (which is a generator of $$\pi_2(S^2)=Z$$)to the class of the projection map [\pi:S^2\rightarrow RP^2].

So the map that like you said is "the identity map to map s2 to s2 and then project to rp2", which is just the projection map pi:S²-->RP² is in fact a generator of $$\pi_2(RP^2)=Z$$.

 

[wdlang]

So now it seems like you're talking about finding a map f:S²-->RP² that is a generator of $$\pi_2(RP^2)=Z$$.

But that is easy given the usually way of determing $$\pi_2(RP^2)$$.

Namely, under the projection (or quotient) map pi:S²-->RP², S² is a fiber bundle over RP^2 of fiber S^0={-1,1}. So there is a long exact sequence of homotopy groups which around pi_2 looks like:

$$0=\pi_2(S^0)\rightarrow \pi_2(S^2)\stackrel{\pi_*}{\rightarrow} \pi_2(RP^2)\rightarrow \pi_1(S^0)=0$$

So $$\pi_*$$ is an isomorphism. And what does it do? It takes the class of the identity map [id:S^2\rightarrow S^2] (which is a generator of $$\pi_2(S^2)=Z$$)to the class of the projection map [\pi:S^2\rightarrow RP^2].

So the map that like you said is "the identity map to map s2 to s2 and then project to rp2", which is just the projection map pi:S²-->RP² is in fact a generator of $$\pi_2(RP^2)=Z$$.

it seems that you have a profound awesome understanding of the subject

i will think of it