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Let $M$ be a smooth manifold and $\pi:TM\rightarrow M$ be the projection map. Define $V=ker(d\pi)$ and define $H$ to satisfy $TTM=V\oplus H$ and suppose $H$ is invariant under the tangent map induced by the multiplication maps $m_\lambda:TM\rightarrow TM:x\rightarrow \lambda x$.

Let $\rho_V:TTM\rightarrow ker(d\pi)$ be the vertical projection map.

Define $\bigtriangledown_X Y:=\rho_V\circ dY \circ X$ for smooth vector fields $X,Y:M\rightarrow TM$.

I was trying to prove that it satisfies that $\bigtriangledown_X (fY)= X(f) Y + f\bigtriangledown_X Y$ where $f$ is a smooth function, but I don't know how to prove this. How do I prove this?

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    $\begingroup$ Although there is already a flawless answer in local coordinates by @Anthony Carapetis, you might profit from taking a look at Lee's Manifolds and Differential Geometry ch. 12, specially the theorem 12.32 that gives a coordinate-free proof of the properties for a connection using the connector map .It suits very well the language you're using. $\endgroup$ Feb 21 '18 at 0:50
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Fix local coordinates $x^i$ on $M$ and define the usual coordinates $(x^i, \xi^i = dx^i)$ on $TM$. The vertical bundle is then just the span of $\partial/\partial \xi^i$, while the horizontal space at $v \in T_p M$ can be written as the span of the vectors $\def\pfrac#1#2{\frac{\partial #1}{\partial #2}}$ $$h_i=\pfrac{}{x^i} - \Gamma_i^j(v)\pfrac{}{\xi^j}$$ for some coefficients $\Gamma_i^j(v)$. You can check that we then have the formula $$\nabla_X Y = X^i \left(\pfrac{Y^j}{x^i}+\Gamma^j_i(Y)\right)\pfrac{}{x^j}\tag{1}$$ for the covariant derivative. When $\Gamma^j_i$ is linear (i.e. $\Gamma^j_i(Y) = \Gamma^j_{ik} Y^k$) this is exactly the usual coordinate formula for an affine connection with Christoffel symbols $\Gamma^j_{ik}.$ The assumption that $H$ is $m_\lambda$-invariant is not quite as strong as linearity, but it does imply positive homogeneity: in coordinates we have $m_\lambda(x,\xi) = (x, \lambda \xi)$; so $Dm_\lambda(\partial/\partial x^i) = \partial/\partial x^i$ and $Dm_\lambda(\partial/\partial \xi^i) = \lambda \partial/\partial \xi^i$ and thus $$Dm_\lambda(h_i|_v) = \pfrac{}{x^i}\Big|_{\lambda v} - \lambda \Gamma^j_i(v) \pfrac{}{\xi^j}\Big|_{\lambda v}$$which can only equal $h_i|_{\lambda v}$ if we have $\lambda \Gamma^j_i(v) = \Gamma^j_i(\lambda v).$ (edit: actually, homogeneity is just as strong as linearity if you're assuming differentiability - see here.)

We now have everything we need to prove the Leibniz rule: applying $(1)$ to both $Y$ and $fY$ we find $$\begin{align} (\nabla_i Y)^j &= \pfrac{Y^j}{x^i}+\Gamma^j_i(Y) \\ (\nabla_i (fY))^j &= f \pfrac{Y^j}{x^i} + \pfrac{f}{x^i} Y^j + \Gamma^j_i(f Y). \end{align}$$

Applying the homogeneity of $\Gamma$ to the last term and multiplying the first equation by $f$ we see $$(\nabla_i(fY))^j = f(\nabla_i Y)^j + \pfrac{f}{x^i} Y^j$$ as desired.

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