Choose a notion of an "ambient homotopy" between maps of topological spaces. For example, say that two embeddings $Y \rightarrow X$ are ambiently homotopic if there is a path between them in the space of embeddings from $Y$ to $X$.


Q. Given a fiber bundle $p: E \rightarrow B$, and two ambiently homotopic maps $f, g : X \rightarrow B$, are pullbacks $X\_f \rightarrow E$ and $X\_g \rightarrow E$ of $f$ and $g$ along $p$ ambiently homotopic (modulo somehow identifying $X\_f$ with $X\_g$)?


If one takes $X$ to be the space with two disjoint points, then $X\_f$ and $X\_g$ both are two disjoint copies of the fiber of $p$, so they are homeomorphic. Using some homeomorphism to identify $X\_f$ and $X\_g$ and using the above definition of the ambient homotopy, what is the answer to the question? 


For example, consider the Hopf bundle $S^3 \rightarrow S^2$. Take $X$ to be the two point set. Then the two pullpacks $X\_f \rightarrow S^3$ and $X\_g \rightarrow S^3$ will both be two linked circles in $S^3$. These are "ambiently homotopic".


More generally, how to define the appropriate notion of an "ambient homotopy" so that the question makes sense?



"Ambient homotopy" between preimages under a fiber bundle?
The comment above was a bit rushed, sorry. Let $H \colon X \times [0,1] \to B$ be a homotopy between $f$ and $g$ and consider the pullback $(X \times [0,1])\_{H}$ of the bundle $p$ along $H$. The pullback comes with a map 
$$ (X\times[0,1])\_{H} \xrightarrow{\phi} E $$
covering the homotopy $H$. If for each $t$, the map $x \mapsto H(x,t)$ is an embedding, the restriction of $\phi$ to the bundle over $X\times\{t\}$ is also an embedding. 


In any case, the point is that $(X\times [0,1])\_H$ is isomorphic as a bundle to the product map $X\_f \times [0,1] \rightarrow X \times[0,1]$ (see for instance Theorems 9.6 to 9.8 in [Husemoller - Fibre bundles](http://www.maths.ed.ac.uk/~aar/papers/husemoller), pages 51-52) hence $\phi$ is your desired "ambient homotopy".


A choice of isomorphism between $(X\times[0,1])\_H$ and $X\_f \times [0,1]$ gives you the way of identifying $X\_f$ with $X\_g$.


