# “Ambient homotopy” between preimages under a fiber bundle?

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?

• Something in your question does not parse for me. The two pull-backs are maps $f^*p : E_f \to X$ and $g^*p : E_g \to X$ these are maps from two different spaces to $X$, so your definition of ambient homotopy does not make sense. – Ryan Budney Nov 17 '15 at 0:25
• A priori, the pullbacks of f and g along p have different codomains $f^*E$, $g^*E$. To make sense of the question, one has to identify these somehow; one certainly can do so, since they’re pullbacks of homotopic maps into a bundle, but the answer may be sensitive to how one does so, and how nice your notion of ambient homotopy is. – Peter LeFanu Lumsdaine Nov 17 '15 at 0:26
• @RyanBudney: if $f, g$ are supposed to be embeddings, then that rules out the constant map. – Peter LeFanu Lumsdaine Nov 17 '15 at 1:31
• There is a notion of when maps $f: Y\to X$ and $f': Y'\to X$ are homotopically equivalent. This means there is a homotopy equivalence $h: Y\to Y'$ making the evident triangle commute. I think that is the best you can hope for here. – Mark Grant Nov 17 '15 at 10:06
• Now that the question has been changed, the answer is yes. If $f_t$ is an isotopy between $f$ and $g$ (say $f_0=f, f_1=g$) then $X_{f_t}$ will embed for each $t$. By homotopy invariance of bundles, for each $t$, $X_{f_t}$ is equivalent to $X_f$ as a bundle over $X$. The choice of identification is a choice of isomorphism between the pullback via the isotopy and the bundle $X_f \times[0,1]$. – Gustavo Granja Nov 17 '15 at 16:28

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, 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 $X_g$.