vector bundles on $X \times S^2$ built via clutching functions

Question

Let $X$ be a compact Hausdorff space and $E \to X$ a (complex) vector bundle over $X$. We can build a vector bundle on $X \times S^2$ by clutching functions, i.e an automorphism $f:E \times S^1 \to E \times S^1$ of vector bundle over $X \times S^1$, then the associated bundle $[E,f]$ on $X \times S^2$ is the quotient

$[E,f] := E\times D^2 \sqcup E \times D^2$ with the identification $(v,x) \sim f(v,x)$

where the $D^2$'s are the upper and lower hemispheres of the $S^2$ and they intersect in the equatorial $S^1$.

On p.43 (see excerpt below) of Allen Hatcher's K-theory book it says any clutching function $f$ can be normalized, (I assume without affecting the isomorphism class of $[E,f]$), and any two normalized clutching functions are homotopic.

Since homotopic clutching functions give rise to isomorphic vector bundles, this seems to suggest all vector bundles of the form $[E,f]$ for fixed $E$ but variable $f$ are isomorphic. This doesn't seem right. Where am I going wrong? Here is the relevant paragraph from Hatcher:

EDIT: Oh, I think I see my misunderstanding of what Hatcher is trying say. He is saying that if we start with a vector bundle $E'$ on $X \times S^2$, then we can associate to it a normalized clutching function $f$, which is unique up to homotopy, such that $E' \approx [E,f]$. If $f_0, f_1$ are two such normalized clutching functions that we get by starting off with the same bundle $E'$ on $X \times S^2$, then they are homotopic (and so then of course $[E, f_0] \approx [E,f_1]$ since they are both isomorphic to $E'$, the bundle we began with). I didn't realize he was talking about starting with the same bundle $E'$.

Answer 1

I'm no expert on the matter and this was intended as a comment but it ended up to be too long.

I think what is meant in the last sentence of the first paragraph you cited is that $h_0h_{\infty}^{-1}$ induces a clutching function $f$ for $E'$ by restricting to $E\times S^1$.

The last sentence in the second paragraph seems a bit strange to me...
As far as I can tell, what's going on is the following:
Any two normalized (in the sense Hatcher describes) automorphisms $f_0,f_1\in Aut(E\times D)$ of the product bundle $p\times id :E\times D\to X\times D$ are homotopic through normalized automorphisms $f_t\in Aut(E\times D)$. But I see no reason why their respective restrictions to $E\times S^1$ (i.e. the clutching functions they induce) should be homotopic through automorphisms of the bundle $E\times S^1$. So, indeed, not all vector bundles of the form $[E,f]$ for fixed $E$ and variable $f$ are isomorphic.

Answer 2

You can normalize every automophism in this way:

If we restrict the isomorphism $h_a \colon E_a \to E \times D_a$ to the part of $E_a$ over $X \times \{1\}$, we obtain an isomorphism $E \to E \times \{1\}$ sending a vector $v \in E$ to a pair $(g(v),1)$ in $E \times \{1\}$, where $g$ is an isomorphism $E \to E$. Now compose $h_a$ with the isomorphism $E \times D_a \to E \times D_a$ sending $(v,z) \mapsto (g^{-1}(v),z)$

Every two choices of normalized $h_{\alpha}$ are homotopically equivalent, throughout $h_{\alpha}$'s normalized clutching functions. In fact every two functions differ for a map $g_{\alpha} \colon D_{\alpha} \to Aut(E)$ with $g_{\alpha}(1)=\text{Id}_{Aut(E)}$ (because the two functions are normalized over $1$). Such $g_{\alpha}$ is homotopically equivalent to the Identity thanks to the deformations retract of the disk to the point $1$. So we obtain our NORMALIZED homotopy (deformation retract leaves the point fixed!).

yes your EDIT is correct.