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How to solve this?

Let $f\colon M \to N$ and $g\colon N\to L$ be smooth maps.

  1. Show that if $f$ and $g$ are embeddings, then $g\circ f$ is an embedding.
  2. Show that if $g$ and $g\circ f$ are embeddings, then $f$ is an embedding.
  3. Find an example where $f$ and $g\circ f$ are embeddings but $g$ is not an embedding. (You can find such an example where $M$, $N$, and $L$ are open subsets of $\mathbb{R}$).

Edit:
Can only use this definition of an embedding:
A smooth map f:N->M is an embedding if
(I) It is a one-to-one immersion and
(II) the image f(N) with the subspace topology is homeomorphic to N under f.

I think I have shown (I), for 1. and 2., how to show (II)?

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6  
What have you tried for yourself? –  Asaf Karagila Apr 29 '11 at 9:00
    
There are different notions of embeddings. In differential geometry, an embedding is a smooth map that is a diffeomorphism onto its image, but in topology it need only be continuous. Which one? –  Fredrik Meyer Apr 29 '11 at 13:43

2 Answers 2

up vote 1 down vote accepted

For (1), just try applying the definition. $M$ is homeomorphic to $f(M)$ because $f$ is an embedding, and $f(M)$ is homeomorphic to $g\left( f(M)\right)$ because $g$ is an embedding. Therefore, $M$ is homeomorphic to $g\left( f(M)\right)$.

EDIT: I should mention that we want more than $M$ just being abstractly homeomorphic to $g\left( f(M)\right)$. We want a homoemorphism to be given by $g\circ f$, not just any old homeomorphism floating about. Of course, this argument gives us that, I just thought I should point out this subtle detail that wasn't quite so obvious from the way I worded my post.

For (2), try composing $g\circ f$ with $g^{-1}$. Note that, $g^{-1}:\mathrm{Im}[g]\rightarrow N$ is an embedding because $g$ is. Then, just apply a similar argument as you did in (1).

For (3), try $M=(0,1)$, $N=L=\mathbb{R}$, $f(x)=x$, and $g(x)=x^2$. If you try a similar trick here as you did in (2), you find that $(g\circ f)\circ f^{-1}=g|_{\mathrm{Im}[f]}$ is an embedding. They key is that $g$ restricited to the image of $f$ works just fine, but that's not what we care about. We care about $g$ being an embedding on the entire domain.

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Hint: Compositions of injective/surjective functions are injective/surjective

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Ok, so is this sufficient for (I) ? –  pi_yum_yum May 1 '11 at 10:15
    
And what theorem is this? Does it need proof? –  pi_yum_yum May 1 '11 at 10:25
    
And you need the chainrule... –  Thomas Rot May 1 '11 at 12:02

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