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I am reading the following definition:

A Lagrangian submanifold $j:\Lambda\hookrightarrow T^*Q$ is transversal to the fibers of $T^*Q$ if given $$\Lambda\hookrightarrow T^*Q\xrightarrow{\pi_Q} Q$$ $$\lambda\mapsto (q(\lambda),p(\lambda)) \mapsto q(\lambda)$$ then $T_{q(\lambda),p(\lambda)}T^* Q = T_{q(\lambda),p(\lambda)}j(\Lambda)\oplus T_{q(\lambda),p(\lambda)} \pi^{-1}(q(\lambda))$.

Then with no further explanation, the statement that this is equivalent to $\hbox{rk}(D(\pi_Q\circ j))=\dim Q$ is made. How to prove that these are equivalent? I have been trying to work with the tangent maps but I am having a rough time proving this since I lack a good background in differential geometry.

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    "Transversal to the fibers" means that at every point of the submanifold the tangent space to the submanifold and the tangent space of the fiber going through that point intersect trivially (and therefore that their sum is the whole tangent space of T^*Q at that point and that the sun is direct) – 2017-02-15
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    To understand the equivalent formulation, I suggest you consider the case where Q is just R^n, or even just R. – 2017-02-15

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