Show that there exists no immersion f of S 1 into R^1 I'm studying Differential Forms and Applications by Manfredo P.do Carmo.
First, I suppose that there exists f s.t. df is injective.
I guess the problem can be solved to use Stoke's theorem and other lemma.

Lemma. 
M : a compact orientable manifold with boundary M = empty.
w : a (n-1)-form on M.
Then there exists a point p in M s.t. dw(p)=0.

But I can't use them exactly so please help me..
 A: I'm not exactly sure what else you're looking for other than what's been provided in the comments, but I hope this helps. In order to use the lemma, it's important to understand that the set of 0-forms $\Omega^0(M)$ on a smooth manifold $M$ is by definition the set of smooth functions $M \to \Bbb{R}$. Thus, as Ivo points out, your $f$ is itself a 0-form on $S^1$. Since $S^1$ is a 1-dimensional manifold, the lemma applies and tells you that there is some point $p \in S^1$ such that $df(p) = 0$, and hence is not injective at this point. 
For a proof not using the lemma, and one that doesn't really rely on differential forms, note that the image of a compact space under a continuous map is again compact. As you've noted, $S^1$ is compact, so $f(S^1) \subseteq \Bbb{R}$ is compact and hence closed and bounded. There is therefore some largest value in $f(S^1)$, in other words there is some point $p \in S^1$ at which $f$ attains a maximum. Now, if you like, pick some coordinate system $(a,b)$ around the point $p$ and call the variable $t$. Then viewing $f: (a,b) \to \Bbb{R}$, $df$ is just $\frac{df}{dt}dt$. Now only using single variable calculus we know that $\frac{df}{dt}=0$ at a maximum of $f$, so $df(p)=0$.
