I'm reading the book "Classical and Modern Morse Theory with Applications" by Mercuri, Piccione and Tausk.
We want to characterize the points $q\in \mathbb{R}^{n+p}$ such that the distance function $L_q(x)=\langle q-f(x), q-f(x) \rangle$ is Morse. Here, $M$ is an $n$-dimensional Riemannian manifold, $L_q : M \to \mathbb{R}$ and $f:M \to \mathbb{R}^{n+p}$ is an isometric immersion.
It was also obtained before that if $\xi (x)= q-f(x)$, then $\overline{\nabla}_X \xi =-X$, so $$dL_q (x) X = -2\langle X, \xi \rangle$$ and $$d^2 L_q (x) (X,Y)=-2Y \langle X, \xi \rangle =2 \langle(Id-A_{\xi})X,Y \rangle,$$ where $A_{\xi} X = - \overline{\nabla} _X \xi$ denotes the shape operator at $x$ on the $\xi$ direction.
Having this in mind, the book continues saying that this will be done in terms of the $\textit{ endpoint map }$ or normal exponential map: $$E:\nu M \to \mathbb{R}^{n+p}, \; \; E(x, \eta)=f(x) + \eta,$$ where $\nu M$ denotes the normal space corresponding to $f$.
And it goes:
We compute the differential of $E$. Let $(x, \eta) \in \nu M $ and $\gamma (t) = (x(t),\eta(t))$ be a curve in $\nu M$ such that $x(0)=x, \; \eta (0)= \eta$. Then: $$dE(x, \eta)(\dot{\gamma}(0))= (x(t)+\eta(t))'(0)=\dot{x}(0)+(\dot{\eta}(0))^T + (\dot{\eta}(0))^{\perp},$$ where, for $z\in T_x \mathbb{R}^{n+p}$, $z^T$ y $z^{\perp}$ denote the projections of $z$ in $T_x M$ and in $\nu _x M$, respectively. In particular, taking $x(t)=x, \; \eta(t)=x+t\eta,$ we get that the differential of $E$ along the fibres is the identity (which was geometrically obvious). In particular, $dE (x, \eta)$ and $(Id-A_{\eta})$ haver kernels of the same dimension.
The part that I still don't understand is: "In particular, taking $x(t)=x, \; \eta(t)=x+t\eta,$ we get that the differential of $E$ along the fibres is the identity (which was geometrically obvious)".
I am confused because I think that if we take the dot product $\langle x(t), \eta (t) \rangle = \langle x , x +t\eta \rangle = \langle x , x \rangle \neq 0$ and this should equal $0$ so that the curve is in $\nu M$, so I don't understand why is it taking that curve.
And even if we take that curve, in what sense is the differential of $E$ "along the fibres" the identity and why is that geometrically obvious?. If you could explain some intuition here it would be great, but I am also confused with how to obtain that the differential of $E$ is the identity, since it is a transformation between the tangent space of the normal bundle at $x$ and $\mathbb{R}^{n+p}$, am I wrong?
The last part "$dE (x, \eta)$ and $(Id-A_{\eta})$ have kernels of the same dimension" sounds like the most important to make the characterization, but I have no clue about how to show that.
I would be really thankful if you could help me with any of these parts!