Constructing the Quotient vector Bundle For a vector bundle $E \longrightarrow X$ with a sub bundle $E' \subset E$, I want to construct a quotient vector bundle $E/E' \longrightarrow X$. 
My thoughts are as follows. Let $\varphi : E \longrightarrow E'$ be the surjection of $E$ onto its sub bundle $E'$. Then by the first isomorphism theorem, $E' \cong E/\ker \varphi$. By the second isomorphism theorem, it follows that $E/E' \cong \ker \varphi$. 
Is $\ker \varphi$ still a vector bundle, and is my construction correct?
 A: You have to be a bit careful as there is no "canonical" surjection $\varphi \colon E \rightarrow E'$! 
If $V' \leq V$ is a subspace of a vector space $V$, we don't have a canonical surjective map $\varphi \colon V \rightarrow V'$. Such a map provides a way to complete $V'$ into a direct sum (by setting $V'' = 
\ker(\varphi)$ and then $V = V' \oplus V''$) and there are many ways to do it. 
If you can justify why such a map $\varphi \colon E \rightarrow E'$ exists, then your approach would work and would provide an isomorphism between $\ker \varphi$ and $E / E'$ but this is completely unneccesary if you only want to construct the bundle $E / E'$. Consider an equivalence relation on the total space $E$ given by $v_x \sim v'_y$ if and only if $x = y$ and $v_x - v'_x \in E'_x$ (where we use the notation $v_x$ to denote an element of the fiber $E_x$ of $E \rightarrow X$) and define the obvious projection map $\pi \colon (E / \sim) \rightarrow X$. Then show that $E / \sim$ is a vector bundle over $X$ whose fiber is naturally isomorphic to $E_x / E'_x$ and so deserves the name $E / E'$.
