isomorphic pullback through projections determine isomorphic vector bundles The context is the following: working with $\xi = (E,p_E,X),\eta = (F,p_F,X)$ two vector bundles, denote $\pi_0:X \times D_0 \to X,\pi_{\infty} : X \times D_{\infty} \to X$ the projections over the first factor, where $D_0 := \{z \in \mathbb{C} | |z| \leq 1\},D_{\infty} := \{z \in \mathbb{C} | |z| \geq 1\} \cup \{\infty\}$ so that the clutching costruction, denoted by $[\xi,u]$ is defined for some $u$. In general is it true that if $[\xi,u] \simeq [\eta,u]$ as vector bundles over $X \times S^2$ then $\eta \simeq \xi$ as vector bundles to $X?$ It seems to me due to the fact that since $u$ is the same this implies that in particular $\pi_0(\xi) \simeq \pi_0(\eta), \pi_{\infty}(\xi) \simeq \pi_{\infty}(\eta)$. Since $\pi_0$ and $\pi_{\infty}$ are projections, an isomorphism of two such that vector bundles determines an isomorphism $w : \xi \to \eta$.
Is my intuition correct or otherwise, what am I missing? Any help would be appreciated.
Edit : The question could be rephrase providing less content as follows, is it true that if $\xi = (E,p_E,X)$ is a vector bundle and $f : B\times X \to X$ is the projection on the second factor then it holds $$f^*(\xi) \simeq (E \times B, p_E\times id, X \times B)$$ and and if $\eta = (F,p_F,X)$ is another vector bundle an isomorphism between $f^*(\xi) \simeq f^*(\eta)$, this determines an isomorphism of $\xi \simeq \eta$?
 A: This is a partial answer to the last part of your question, giving the general setting. Your particular case on $X\times S^2$ could have particularities I am not aware of.
The pullback bundle has a very explicit expression:
$$
f^*(\xi) = \{ ((x,b),e))\in (X\times B)\times E | f(b,x) = p_E(e)\}
$$
but the latter condition is equivalent to $x=p_E(e)$, i.e. $e$ must be in the fiber of $E$ above $x\in X$, $E_x$.
Bear in mind that the pullback bundle comes with the projection
$$
((x,b),e)) \mapsto (x,b)\in X\times B
$$
Then, the map
$$
f^*(\xi) \ni ((x,b),e) \mapsto (e,b)\in E\times B
$$
can be seen to be a vector bundle isomorphism covering the identity in $X\times B$. This will be your desired isomorphism.
The pullback construction is said to be natural, in categorical terms, so isomorphic vector bundles over $X$ are pulled back to isomorphic vector bundles over $X\times B$. Actually more is true: a vector bundle map between vector bundles $E_1 \rightarrow E_2$ over $X$ transforms to a map between the corresponding pullbacks, sitting in a nice commutative diagram.
Isomorphism between pullbacks does not in general implies isomorphism between the original bundles. As a counterexample consider this question. There, it is shown that the pullback of the tautological bundle over $\mathbb{CP}^n$ to $\mathbb{C}^{n+1}\setminus\{0\}$ is trivial. However, the trivial bundle over $\mathbb{CP}^n$ will also be pulled back to the trivial bundle: pullbacks are isomorphic, while the original bundles are not.
