I am struggling with page 36 of "Topology from a differetialble point of view" (University press of Virginia) by Milnor, in particular with the proof of Hopf's Lemma. I'm giving the details of the part of the proof I don't understand. Let $M$ be a m-dimensional oriented manifold, with boundary, and let $\bar{v}:\partial M \to S^{m-1}$ be a smooth vector field so that $\bar{v}(x)$ is a pointing-out-vector for each $x\in \partial M$. Let also $g:\partial M \to S^{m-1}$ be the gaussian map (pointing outward) of $\partial M$. Now I have two questions:
- how is it possible to define explicitly a smooth homotopy between $\bar{v}$ and $g$?
- Where the hypothesis that both the fields point outward is crucial in the determination of the homotopy?
You forgot to explain that $M\subset\Bbb R^m$ in the first place so that this makes sense.
Milnor has normalized his vector field, so that $\bar v$ is a unit outward-pointing vector field on $\partial M$. $g$, the Gauss map of $\partial M$, is likewise a unit outward-pointing vector field. You can do the usual homotopy: Join the two by the straight-line path and then collapse back to the unit sphere. That is, take $$H(x,t) = \frac{(1-t)\bar v(x) + tg(x)}{\|(1-t)\bar v(x) + tg(x)\|}.$$ Because both vector fields are outward-pointing, one can never be a negative multiple of the other and this map is (smooth and) well-defined.
This proof will answer your second question. You don't need the vector fields both to be literally outward-pointing, but you do need to make sure they are never opposite one another.