Continuous bijection that is not an embedding

1. Apr 25, 2010

Bacle

Hi:
Just curious: a continuous function f:X-->Y ; X,Y topological spaces, can fail
to be an embedding because it is not 1-1, or, if f is 1-1 , f can fail to be an
embedding because, for U open in X f(U) is not open in f(X).

Can anyone think of a "reasonable" example of the last case, by reasonable
I mean no spaces with finitely many points, the inclusion map from a space
into itself with a different topology (i.e.: we have (X,T) and (X,T') , and we
use i:X-->X : i(x)=x ) , or maybe or something one could find in
"Counterexamples" book.

If A is a strict subspace of X , then the subspace topology on A guarantees
that i:A-->X , the inclusion, is an embedding. Is this true for other topologies on A.?

Thanks.

2. Apr 25, 2010

g_edgar

X = interval [0,2 pi), Y = circle {(x,y) : x^2+y^2=1}, continuous bijection f(t) = (cos(t),sin(t)).

3. Apr 25, 2010

g_edgar

>> If A is a strict subspace of X , then the subspace topology on A guarantees
that i:A-->X , the inclusion, is an embedding. Is this true for other topologies on A.?

No. In fact, that is the definition of "subspace topology" ... so for any topology on A other than the subspace topology, the inclusion i is NOT an embedding. Either because i is not continuous, or because i is not open, or both.

4. Apr 25, 2010

Bacle

Edgar wrote
" No. In fact, that is the definition of "subspace topology" ... so for any topology on A other than the subspace topology, the inclusion i is NOT an embedding. Either because i is not continuous, or because i is not open, or both. "

Thanks, Edgar. I thought the subspace topology was the initial topology with respect
to inclusion, and, given an inclusion:

i:A-->X

the initial topology was (is) , by def., the smallest topology on A, which makes the
inclusion continuous (largest topology being the discrete one, 2A), but
I don't see how it follows that the subspace topology is the only one for which i is
( I am.? :) ) an embedding:

We have A< X a strict subspace ( i.e, A is not all of X ). Given the map : i:A-->X

We want to define a topology on A such that :

1) Continuity of i: For U open in X , i-1(U) open in A.

We have that i-1(U) =U/\A . So we must have U/\A open in A.

Then any topology TA on A must contain the subspace topology.

i.e., TA> (A, subspace) (with > meaning contains)

2) Openness of map i: For V open in A, i(V)=V is open in X . Like you said,
it follows that A must be an open subspace of X, i.e., A is open as a subset
of X.

How does it then follow that the only topology that makes i:A-->X
into an embedding is (A, subspace) .?. I don't see how we can conclude,
e.g.,
(A, subspace)> TA

5. May 24, 2010

Landau

Instead of No. In fact, that is the definition of "subspace topology"', I think g_edgar should have written No. In fact, that is the definition of "topological embedding"'.

Let X,Y be topological spaces. A function f:X->Y is an embedding if
* f is injective, and
* f is a homeomorphism onto its image f(X), where f(X) carries the subspace topology of X.

So the requirement of "subspace topology" is part of the definition of embedding.

6. May 25, 2010

RedX

Doesn't your second *, that f is a homeomorphism onto its image f(X), automatically mean that f is injective, your first *?

7. May 25, 2010

Landau

Yes, you are right, the first * is superfluous. So, in short, an embedding is a homemomorphism onto its image.

8. May 25, 2010

RedX

That makes really good intuitive sense. But for some reason the book I have defines embedding in a slightly esoteric way. Instead of topological spaces, it talks about embedding of differentiable manifolds.

First it defines an immersion. Basically, a smooth map f: M -> N between manifolds induces a map called the "differential map" f* between the vector spaces of the two manifolds:

f*: T(M) -> T(N) (explicitly, this map happens to be the Jacobian).

If f* is an injection, then f is said to be an immersion.

In addition, if f is an injection, then f is said to be an embedding.

This definition of an embedding seems weird to me, not as nice as your definition. Is there a reason why one would define an embedding in this way?

9. May 25, 2010

Landau

Remember, we were talking about topological spaces, i.e. spaces with as additional structure (only) a topology. My definition was of a topological embedding, which is just called an embedding if the context (namely, topology only) should be clear.

In your book it is about (smooth) manifolds: those are first of all topological spaces, but they also have extra structure, namely a smooth structure (atlas). An embedding between smooth manifolds, let us call it a smooth embedding to be sure, also has to deal with this smooth structure, and that is the immersion part.

So, a smooth embedding is a topological embedding which is at the same time an immersion (= injective differential).