Properly embedded submanifold

[JYM]

I try to solve the following problem: If S be submanifold of M and every smooth function f on S has a smooth extentsion to all of M, then S is properly embedded. [smooth means C-infinity].
I can show that S is embedded. What I need is to show either S is closed in M or the inclusion map is proper. If you have any suggestion please well come. Thank you in advance.

 

[fresh_42]

By which condition is $S=M$ ruled out?

 

[lavinia]

I try to solve the following problem: If S be submanifold of M and every smooth function f on S has a smooth extentsion to all of M, then S is properly embedded. [smooth means C-infinity].
I can show that S is embedded. What I need is to show either S is closed in M or the inclusion map is proper. If you have any suggestion please well come. Thank you in advance.

I am not sure what your question is. If you already know that the manifold is embedded: that is its topology is the same as its subspace topology in M then it is homeomorphic to its image in M and a homeomorphism is a closed mapping.

 

[mathwonk]

well, for starters, suppose S is not closed in M and there is a point p of M that is a limit point of S. Can you construct a smooth function f on M that is zero say only at p? what if you then take 1/f? what would that mean?

 

[JYM]

Oh, thanks, I see it. 1/f cannot be extended to all of M since f is zero at p.
I see the fact that S is embedded follows from the following fact but I can't justify.
Let M be a manifold and ϕ : S → M be an injective immersion. Show that ϕ is an embedding if and only if every smooth function f : S → R has an extension to a neighborhood U of ϕ(S).

 

[JYM]

I am not sure what your question is. If you already know that the manifold is embedded: that is its topology is the same as its subspace topology in M then it is homeomorphic to its image in M and a homeomorphism is a closed mapping.

Lavinia, A homeomorphism is not necessarly closed.

 

[lavinia]

Lavinia, A homeomorphism is not necessarly closed.

A homeomorphism is always a closed mapping - the images of closed sets are closed.( Also the images of open sets are open.)

You said that you knew that the manifold was "embedded" which I took to mean that the image of the mapping of $S$ into $M$ was homeomorphic to $S$ in the subspace topology (which you can show that using the function extension property) .

- Here is classic example of an immersed submanifold that is not homeomorphic to its image in the subspace topology.

Think of a torus as the quotient of the Eulicdean plane obtained by identifying points whose $(x,y)$ coordinates differ by an integer. A straight line that makes an irrational angle with the $x$-axis projects to an immersed 1 dimensional submanifold.

The image of the line is dense in the torus - meaning that any point in the torus is the limit of a Cauchy sequence of points on the line. So there is no way to isolate the image of an open interval through an intersection with an open set in the torus.

 

[mathwonk]

what about the inclusion of the open disc into the plane? is the image homeomorphic to the source? is the map closed? i think the disagreement is over considering the map as a map into its image versus as a map into the larger ambient manifold.

 

[lavinia]

I see your point. The mapping is closed in the subspace topology but if the image isn't closed then it can not map the manifold into a closed set.

So it seems that the function extension property gives the image of $S$ as homeomorphic to $S$ i.e. it is an embedding and if the image is also closed in $M$ then the embedding is a closed mapping.

 

[mathwonk]

there are also subtleties to the extension properties. if S is any submanifold of M (such as the disc in the plane) then every smooth function on S extends smoothly to a neighborhood in M of any given point p of S, but if every smooth function on S extends globally to a smooth function on M, then S must be a closed submanifold of M.

 

[lavinia]

there are also subtleties to the extension properties. if S is any submanifold of M (such as the disc in the plane) then every smooth function on S extends smoothly to a neighborhood in M of any given point p of S, but if every smooth function on S extends globally to a smooth function on M, then S must be a closed submanifold of M.

@mathwonk To see if I understand your points:

- If a function on a submanifold diverges then it cannot be extended to the entire ambient manifold. Such a function would exist If the submanifold had a boundary point. So the function extension property implies that submanifold must be closed.

- If the submanifold is a closed subset then the immersion is a closed mapping. One does not need to worry about its subspace topology.

- But it turns out that the function extension property also implies that the subspace topology makes the immersion into a homeomorphism.

- Going the other way if an injective immersion is closed then it might not be an embedding. If the manifold is not compact an immersion of it might turn back on itself such as bending the positive x-axis in the plane around so that as points approach positive infinity the immersion converges to the origin. In this case, the image of the immersion is closed but it is not an embedding.

I think using a tubular neighborhood argument it can be shown that if the manifold is compact then an injective the immersion of it is an embedding.

nu?

 

[mathwonk]

sounds persuasive. i don't think tubular nbhds are needed for your result. if the manifold is compact then the image is closed, and in a compact (Hausdorff) manifold closed sets are the same as compact sets. thus the immersion is both a closed map and continuous onto its image, hence a homeomorphism onto the image, so the example you gave where the image is closed but the map is not closed does not occur. (In your example the unbounded closed set consisting af all reals greater than or equal to some large number, maps to a bounded non closed set approaching the origin, but not containing it.) so in your example the image of the immersion is closed but the map still is not closed. i am assuming your example is essentially the famous injective immersion of the real line onto the closed figure eight, not a closed map and not a homeomorphism onto its (closed) image.

 

[WWGD]

I think a standard could buynot defending is 1/x on (0,1)

 

[lavinia]

I didn't understand your original question. I thought that "closed" meant closed in the subspace topology. If the immersion is an embedding then this is true by definition.

Here are a couple of ideas. The first aims to show that the immersion is an embedding, the second that it is a closed mapping.

Notation: $i:S→M$ is a smooth injective immersion of a manifold $S$ in a manifold $M$. $i(S)$ is the submanifold of $M$.

- Embedding:

If you extend a bump function with support $U⊂i(S)$ to all of $M$ then the inverse image of the positive reals is an open set in the ambient manifold whose intersection with the immersed manifold is $U$.

-Closed Mapping:

I think the function extension property can be used to show that if an infinite sequence in $S$ has no limit point then its image in $i(S)$ can not have a limit point either. Try using bump functions again, this time normalizing them to always achieve the same maximum value.

For this argument it seems necessary to first show that $i^{-1}$ is continuous in the sense that Cauchy sequences in $i(S)$ that converge to a point in $i(S)$ come from convergent Cauchy sequences in $S$.