Horizontal Subbundle I just read the definition of the horizontal subbundle and I have some doubts.
Suppose we endow $M$ with a Riemannian metric . We will define the connection map $K: TTM\rightarrow TM $.Let $\eta \in T_{\theta}TM$ and $z:(-\epsilon,\epsilon):\rightarrow TM$ a curve adpated to $\eta$. Such a curve gives rise to a curve $\alpha(t)=\pi\circ z(t) : (-\epsilon,\epsilon)\rightarrow M$ and a vector field $Z$ along $\alpha$, equivalently , $z(t)=(\alpha(t),Z(t))$.
Now define $K_{\theta}(\eta):=\nabla_{\alpha(t)}Z(t)(0)$. This will be well-defined and linear. The well definiteness I can understand that might come the picard-lindelof theorem and so it is independent of the curve we choose . Now for the linear part I have no idea how to see it , I have tried using local coordinates but got nowhere. Any help is aprecciated . Thanks .
New edit : Ok it's not needed I have used local coordinates again and was able to check it.
 A: Express the connector in a geodesic frame, then it becomes obvious. Here are some more details:
Suppose $(x,v)\in TM$ and $X^1,\dots,X^n$ are smooth vector fields on $M$, defined in a neighbourhood $U$ of $x$. We say that that $(X^1,\dots,X^n)$ is a frame, if $\{X^1(x'),\dots,X^n(x')\}$ is a basis of $T_{x'}M$ for all $x'\in U$ and we say it is a geodesic frame, if additionally $\nabla X^j(x)=0$ for $j=0,\dots,n$. It's a standard excercise to show that such a geodesic frame exists around every point $x$.
Once a local frame is chosen, we can set up an isomorphism $J:T_{(x,v)}TM\rightarrow \mathbb{R}^n\times \mathbb{R}^n$ as follows: If a vector $\xi \in T_{x,v}TM$ is reprensented by the pair $(\gamma,Z)$ (where $\gamma$ is a curve in $M$ and $Z$ is vector field along $\gamma$ such that $\gamma(0)=x$ and $Z(0)=v$), then $\dot \gamma(t)=\sum_{j=1}^n a_j(t)X^j(\gamma(t))$ and $Z(t)=\sum_{j=1}^n b_j(t)X^j(\gamma(t))$ and we define $$J\xi=(a_1(0),\dots,a_n(0),\dot b_1(0),\dots,\dot b_n(0))\in \mathbb{R}^n\times\mathbb{R}^n$$
Now the crux is that $K\xi = \nabla_{\dot\gamma(t)}Z\vert_{t=0}=\nabla_{\dot\gamma(t)} \sum_{j=1}^n b_j(t)X^j(\gamma(t))\vert_{t=0} = \sum_{j=1}^n\dot b_j(0) X_j(x)$, as no derivatives fall on the frame! That means  that in this frame, the connector simply corresponds to the projection onto the second factor, as summarised in the commutative diagram below:
$$\require{AMScd} \begin{CD}
T_{(x,v)}TM @>{K}>>  T_xM\\
@VVV @VVV \\
\mathbb{R}^n\times\mathbb{R}^n@>{\mathrm{pr}_2}>> \mathbb{R}^n
\end{CD}$$
Here the vertical arrows are the natural identifications that come from the frame (i.e. the left one equals $J$). From this point of view, linearity is obvious.
