# Submersion btw R and S^1

1. Oct 26, 2007

### quasar987

I think it's accepted to post HW type question in here.

"Is there a submersion from S^1 to R? From R to S^1?"

By a submersion from M to N, I mean a map f:M-->N whose tangent map is surjective.

I answered 'yes' to both questions, which I find dubious.

Take for an atlas of S^1, {({$\exp(i\theta):0<\theta<2\pi$}, z-->arg(z)), ({$\exp(i\theta):-\pi<\theta<\pi$}, z-->arg(z))}

Submersion from S^1 to R: Let x be in S^1 and f:S^1-->[0,2pi[ be f(z)=arg(z). Let r be a real number. I must show that there is a path y:]-e,e[-->S^1 such that y(0)=x and d/dt(f o y)(0)=r. Well such a path is y(t)=xexp(irt).

Submersion from R to S^1: Let x to be in R and f:R-->S^1 be f(y)=exp(ig(y)), where g:R-->[0,2pi[ is the "mod 2 pi" map. Let r be a real number. I must show that there is a path y:]-e,e[-->R such that y(0)=x and d/dt(p o f o y)(0)=r, for p a chart of S^1 around f(x). Well such a path is y(t)=x(t+1)^(r/x). I have constructed y so that y'(0)=r.

Yes, because by construction, p o f = g, and d/dt(g o y)(0)=g'(y(0))*y'(0)=1*r=r.

2. Oct 27, 2007

### Reverie

What you wrote is wrong and a bit difficult to read.

1.)A submersion from S^1 -> R does not exist.
2.)A submersion from R -> S^1 does exist.

"By a submersion from M to N, I mean a map f:M-->N whose tangent map is surjective."

is missing a key element. Usually a submersion is required to be smooth in addition to having the surjectivity condition you mentioned.

4.)The argument map from S^1 -> R is not continuous. That is the problem with your first proof.

To write more clearly, I suggest the following.

1.) Open parenthesis are usually written using "(" and ")". Most readers like looking at these symbols as opposed to "]" and "[". Use the standard notation. Nonstandard notation requires a reader to visualize the "standard version". This reduces mental capacity of the reader to accomodate other errors, misunderstandings, and varying notational conventions.

2.)Make it clear what you are doing.

"I must show that there is a path y:]-e,e[-->S^1 such that y(0)=x and d/dt(f o y)(0)=r."

I eventually realized what you were doing with this step. However, it took more effort than it should have given that I was familiar with these concepts and definitions.

Instead of writing, "I must show that..."

Write, "In order to show f is surjective, ..."

This helps the reader immediately know what your goal is with that particular step.

3. Oct 27, 2007

### quasar987

Thanks for correcting my definition.

But how would a proof of non-existence go? Do you have a hint for me?

4. Oct 29, 2007

### Reverie

Let f:S^1-->R be any smooth map and consider Im(f) subset of R. Let m=max Im(f). At any point x in the inverse image of m, the map f will not be a submersion.

Take local coordinates c around the point x. Then,
f compose c:(-e,e)-->R is not a submersion. Hence, f is not a submersion.

The graph of f compose c subset of (-e,e)xR looks like "n". At the point where the tangent line is horizontal, the function f compose c is not a submersion.