Show that $f \circ g$ is an embedding if $f$ and $g$ are How to solve this?

Let $f\colon M \to N$ and $g\colon N\to L$ be smooth maps. 
  
  
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*Show that if $f$  and $g$  are embeddings, then $g\circ f$  is an embedding. 
  
*Show that if $g$  and $g\circ f$  are embeddings, then $f$  is an embedding. 
  
*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)?
 A: 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.
A: Hint: Compositions of injective/surjective functions are injective/surjective
