# Homework Help: An attempted proof of a theorem in elementary differential geometry

1. Jun 22, 2007

### Saketh

1. The problem statement, all variables and given/known data

For any open set $U \subset \mathbb{R}^n$ and any continuous and injective mapping $f : U \rightarrow \mathbb{R}^n$, the image $f(U)$ is open, and $f(U)$ is a homeomorphism.

2. Relevant equations

N/A

3. The attempt at a solution

I am trying to learn how to write proofs, so my logic might be flawed.

Lemma: f is bijective.

Proof: The cardinality of U and $\mathbb{R}^n$ is the same. Because f is one-to-one, all that remains is to prove that every image has a corresponding preimage, which follows directly from the continuity of f.

f(U) is necessarily open because of the definition of continuity.

The inverse of f exists and is continuous because f is bijective and continuous (I feel like that statement is invalid). Therefore, f is homeomorphic.

2. Jun 22, 2007

### ZioX

f(U) is the image of U under f. You need to show that f acts bijectively from U to Im(f)(=f(U)). Since f is injective this trivially follows.

Since f is injective, there exists a functional inverse whose domain is Im(f). This inverse function is continuous.

Therefore, U and f(U) are homeomorphic.

3. Jun 22, 2007

### matt grime

It is certainly not true that the image of an open set under a continuous map is open. A function is continuous if the *inverse* image of open sets is open.

1. Construct a counter example to your assertion that the image of an open set is open (R to R is easy, with U=(-1,1) so do that).
2. Ask yourself why injective is in the hypothesis.

4. Jun 22, 2007

### Saketh

Okay, thanks for pointing out my errors. Is this correct now?

Lemma 1: f is bijective.

Proof of lemma 1: U and $\mathbb{R}^n$ have the same cardinality, therefore, since we are told that f is injective, it follows that f is also bijective.

Lemma 2: f has a continuous inverse.

Proof of lemma 2: Because f is bijective (lemma 1), its inverse is bijective. Now to prove continuity. Assume that the inverse is discontinuous. This implies that there exists at least one open set in the preimage that has no image. But this would mean that f is not bijective, which violates lemma 1. Therefore, the inverse function must be continuous (is this logic correct?).

From lemmas 1 and 2, it follows that f is a homeomorphism.

5. Jun 22, 2007

### matt grime

The cardinality of U or R^n is immaterial. Any map is surjective onto its image, hence an injection is a bijection between the domain and the range.

Your proof of lemma 2 is wrong. Any set in U must have an image in R^n under f. It is up to you to show that image of an open set is open. Since you've not attempted to do this (and that is the definition of continuous), you should have the warning sirens going off in your head.

6. Jun 22, 2007

### Saketh

Thanks for correcting me again. Let me try once more.

Lemma 2: If has a continuous inverse.

Proof of lemma 2: Because the inverse of f is bijective, every set in the image has one corresponding open preimage in U. Since the total preimage of f consists only of open sets, and f is continuous, the image too consists only of open sets. Thus, the image of an open set U under f must be open.

Because f is an open, continuous, bijective map, it is a homeomorphism.

7. Jun 23, 2007

### matt grime

This is true irrespective of whether f is bijective or not.

I have no idea what you mean by 'total preimage', and can only deduce you're assuming the answer.

this doesn't follow

Again, this doesn't follow at all.

Suppose that V is an open set in U, you need to show that f(V) is open.

You appear to be taking f(V) open and showing V is open, which doesn't do anything.

8. Jun 23, 2007

### Saketh

I'm trying to show that all elements in the domain (total preimage) are open, and that all elements in the image are open. From the bijectivity of f, it follows that for every U, f(U) is open. What is wrong with this?

9. Jun 23, 2007

### matt grime

What do you mean by elements in the domain, or image? To me that can only mean points in U. If you're attempting to describe subsets, then it is obviously not true that the only subsets are open subsets.

Take an open set V<U, and attempt to show that f(V) is open, please.