Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{R}$).

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)?

share|cite|improve this question
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
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.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.