In order to prove the Lemmma $1$ at pag $28$ of differential topology :
Lemma : Let $\tilde{M}$ be a compact $(n+1)-$manifold, $M = \partial \tilde{M}$ with induced orientation, $M$ compact and $f : M \longmapsto N$ of class $C^{\infty}$ with dim$M$=dim$N$. If $f$ extends to some $F : \tilde{M} \longmapsto N$ then deg($f,y) = 0$, for all $y \in \hspace{0.1cm} $RegVal($f$).
We use the following proposition :
$\exists \hspace{0.1cm} X_i : [0,1] \longmapsto T\tilde{M} C^{\infty}$ such that $\forall \hspace{0.1cm} t \in [0,1], (\gamma'(t),X_2(t),\cdots X_{n+1}(t))$ is a positive basis of $T_{\gamma(t)}\tilde{M}$
In particular, the claim is that
Claim :$\text{sgn}df_a +\text{sgn}df_b = 0$
given $A$ an arch of $F^{1}(y)$ where $y \in \text{RegVal}(F)\cap \text{RegVal}(f)$, with $\partial A = \left\lbrace a,b\right\rbrace$, I'm going to link a picture just for the sake of a possible scenario.
I have an issue in the proof of the claim, which goes like this :
Given $\gamma : [0,1] \longmapsto A$, a diffeomorphism with the arc $A$, and taken $X_2,\cdots,X_{n+1} : [0,1] \longmapsto \mathbb{R}^N$ of the proposition, we note that since $F_{|_{A}}$ is costant then $dF_{\gamma(t)}(\gamma'(t)) = 0$ $\forall \hspace{0.1cm} t$.
However, since $dF_{\gamma(t)}$ is surjective, we have that $dF_{\gamma(t)}(X_2(t),\cdots,X_{n+1}(t)) = B_{t}$ is a basis of $T_{y}N$, which has the same orientation at time $t=0,t=1$ since the change of basis in $t=0$ is the identity, and $[0,1]$ is connected, so $B_0 \sim B_1$.
Now, since $y$ is a regular value for $f$, $df_a : T_a M \longmapsto T_y N$ is by dimension injective, hence $\gamma'(0) \not\in T_a M$, and we have that $T_a \tilde{M} = \langle \gamma'(0) \rangle \bigoplus T_a M$.
Let $\pi : T_a M \longmapsto T_y N$ the projection and let's define $Y_{i}(0) := \pi(X_i (0)), i = 2,\cdots,n+1$ be a basis $C_0$ of $T_a M$. Since $Y_i(0) = X_i(0) + \alpha_i \gamma'(0)$, $\alpha_i \in \mathbb{R}$, we have that $df_a(Y_i (0)) = dF_a(Y_i(0)) = dF_a(X_i (0)) \in B_0$, in other words $df_a(C_0) = B_0$. Morover the change of basis from $\gamma'(0),X_2,\cdots,X_{n+1} \to \gamma'(0),Y_2,\cdots,Y_{n+1}$ has positive determinant (the change of basis is a upper triangular matrix with 1 on the diagonal), so those two basis share the same orientation, so the second is a positive basis of $T_a \tilde{M}$.
Now starts the problem : So, since $\gamma'(0)$ points inward, $C_0$ is a negative basis of $T_a M$.
With a similiar process we costruct $C_1$ basis of $T_b M$ with $df_b(C_1) = B_1$ and since $\gamma'(1)$ points outward, $C_1$ is a positive basis of $T_b M$, then in the end $\text{sgn}df_a +\text{sgn}df_b = 0$
My problem is that I don't see how to notice or imply that $\gamma'(0)$ and $\gamma'(1)$ being an-inward or outward vector determines the orientation of $C_0$ and $C_1$ as basis of $T_a M$. I don't see the problem of both pointing outward for example, because I'm not seeing how they have to be forced to be negative and positive while completed with $gamma'(0)$ share the same orientation as the positive basis $\gamma'(0),X_2(0),\cdots,X_{n+1}(0)$ for example .
My definition of outward in a point p, is a vector which lies in $T_p M$ but not in the cone $C_p M$. I thought I could modify a little the curve in a neighborhood of $1$, something like $\alpha(t) = \begin{cases} \gamma(t) \hspace{0.2cm} t \in [1-\epsilon,1] \\ \gamma(1) + (t-1)\gamma'(1) \hspace{0.2cm} t \in [1,1+\epsilon]\end{cases}$ to be in the tangent space but not in the cone; but even if $\alpha$ would satisfy being outward, it seems a choice completely arbitrary and dictated by the example I have in mind and not rigorous.
Any help or explanation will be very appreciated.
