Intersection of subvector bundles Suppose we have a smooth vector bundle $\pi: E \rightarrow B$ and two sub vector bundles $\pi_1: E_1 \rightarrow B_1$ and $\pi_2: E_2 \rightarrow B_2$ such that 
the bases $B_1$ and $B_2$ are submanifolds of $B$. Now suppose we would like to intersect both subbundles, that is we would like to define 'something like' 
$\pi_{1,2}: E_{12} \rightarrow B_1 \cap B_2$ where:
1.) $B_1 \cap B_2$ is a (smooth) submanifold of $B$.
2.) For each $b \in B_1 \cap B_2$ we define 
$\pi_{1,2}^{-1}(b):= \pi_{1}^{-1}(b) \cap \pi_{2}^{-1}(b)$ as the (standard set theoretic) intersection of the fibers of $\pi_1$ and $\pi_2$ and 
$E_{12}:=\bigcup_{b \in B_1 \cap B_2}\pi_{1,2}^{-1}(b)$.
3.) The dimension $dim(\pi_{1,2}^{-1}(b))$ is constant for all $b \in B_1 \cap B_2$.
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Is this a vector bundle?
Is it a sub(vector) bundle of $\pi$?
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P.S.: I know the discussion here: 
About the intersection of two vector bundles
but since it doesn't solve the problem, I think its o.k. to ask my question anyway.
 A: For every  vector space $V$ we have a difference map 
$$ D: V\oplus V\to V,\;\; D(v_0,v_1)=v_1-v_0$$
whose kernel is the diagonal $\Delta_V\subset  V\oplus V$.  More generally, for vector bundles we have a bundle map 
$$D: E\oplus E\to E$$
whose kernel is the diagonal sub-bundle  $\Delta_E$.  Consider now the restriction $\bar{D}$ of $D$ to the subbundle   $F:=E_1\oplus E_2 \subset E\oplus E$. For any $b\in B$ the kernel of $\bar{D}_b$ can be identified with the subspace $E_1(b)\cap E_2(b)$.    This suggests a more general  problem.  

Suppose that $E,F\to B$ are smooth
  vector bundles over a compact smooth
  manifold $B$ and $T: F\to E$ is a
  smooth bundle morphism such that $\dim
> \ker T_b$ is independent of  $b$. Then
  the family of subspaces $ \ker T_b$
  forms  a  smooth vector bundle.

This is certainly the case. To se this equip  $E$ and $F$ with metrics and observe  that
$$\ker T=\ker \left(T^*T: E\to E\right)$$
so we reduce the problem to    the case when $E=F$ and $T$ is selfadjoint and nonnegative definite. This is what I will assume in the sequel. 
For  $b\in B$ we denote by $\lambda(b)$ the smallest  nonzero eigenvalue of $T_b$.   The fact that the dimension of $\ker T_b$  is independent of $b$ implies that
$$\lambda_0:=\inf_{b\in B} \lambda(b) >0. $$
Use the spectral theorem to represent   the projection  onto $\ker T_b$ as a contour integral along a   circle in the plane centered   at $0$ and of radius $\lambda_0/2$. This proves that this projection  depends smoothly on $b$ and thus solves the above problem.
