Equivalence between a vector $v\in T_{p}(M)$ tangent to the fiber $\psi^{-1}(\psi(p))$ and $d\psi(v)=0$ I'm trying to prove the next proposition.
Let $\psi:M\rightarrow B$ be a submersion. Then:
a) A vector $v\in T_{p}(M)$ is tangent to the fiber $\psi^{-1}(\psi v)$ if and only if $d\psi(v)=0.$
b) A map $\phi:B\rightarrow N$ is smooth if and only if $\phi\circ\psi$ is smooth. 
I'm stuck prove them. For part b), I think is enough with the smoothness of $\phi$ and $\psi$ together with theirs charts. 
For part a) I'm stuck. I can't see the connection between the fiber and the diferential $d\psi(v)=0.$
Any kind of help is thanked in advanced.  
 A: a) Since $\psi$ is a submersion, $\psi(v)$ is a regular value of $\psi$ and $\psi^{-1}(\psi(v))$ is a manifold such that:
$$T_v\psi^{-1}(\psi(v))=\ker(\mathrm{d}_{v}\psi).$$
Indeed let $\gamma\colon]-1,1[\rightarrow M$ be a curve drawn on $\psi^{-1}(\psi(v))$ such that $\gamma(0)=v$, then $\psi\circ\gamma=\psi(v)$ so that:
$$0=(\psi\circ\gamma)'(0)=\mathrm{d}_{\gamma(0)}\psi(\gamma'(0))=\mathrm{d}_v\psi(\gamma'(0)),$$
which means $\gamma'(0)\in\ker(\mathrm{d}_v\psi)$ and $T_v\psi^{-1}(\psi(v))\subset\ker(\mathrm{d}_v\psi)$, whence the result looking at dimensions.
b) It is necessary to assume that $\psi$ is surjective, all points of $B$ must be in the image of $\psi$.
Since smoothness is a local property, one may assume freely that $M=\mathbb{R}^m$, $B=\mathbb{R}^b$ and $\psi$ is given by $(x_1,\ldots,x_m)\mapsto(x_1,\ldots,x_b)$, the result is then clear. If the composition $(x_1,\ldots,x_m)\mapsto \phi(x_1,\ldots,x_b)$ is smooth, then the map $(x_1,\ldots,x_b)\mapsto\phi(x_1,\ldots,x_b)$ also is.
