Inverse of spherical fibration I am reading Browder's 'Surgery on simply connected manifolds'. There is a discussion of spherical fibrations in I.4.5 (P.21) that goes roughly as follows:
Fix a connected space $X$. A fibration $E \overset{\pi}{\to} X$ is called a spherical fibration if its fibre is homotopy equivalent to $\mathbb{S}^n$ for some $n$. Any spherical fibration can be embedded as a sub-fibration of a fibration with contractible fibre, e.g. replace $E \overset{\pi}{\to} X$ by the mapping cylinder $M\pi \to X$ (this is similar to a disc bundle for a sphere bundle). Given spherical fibrations $E_1 \overset{\pi_1}{\to} X$ and $E_2 \overset{\pi_2}{\to} X$, embed them as subfibrations of $D_1 \to X$, $D_2 \to X$ which have contractible fibre. The sum of the spherical fibrations $\pi_1+ \pi_2$ is defined to be the spherical fibration $D_1 \times E_2 \cup E_1 \times D_2 \to X$. One can show that the sum is well defined up to fibre homotopy equivalence. 
After this definition, Browder immediately assumes the existence of stable inverses, that is:

For any spherical fibration $E \overset{\pi}{\to} X$, there exists a spherical fibration $E' \overset{\pi'}{\to} X$ such that $\pi+\pi'$ is a trivial fibration.

I cannot see why the stable inverse should exist in general. If the spherical fibration is obtained as the sphere bundle of a vector bundle $V$, then the claim is clear: we can take as the inverse the sphere bundle of $Hom(V,\mathbb{R})$. However for a general fibration, I cannot find an analog for $Hom(V,\mathbb{R})$. I suspect that one might be able to find a H-space $G$ such that the spherical fibrations (modulo fibre homotopy equivalence) are given by $[X,G]$ (the idea inspired from vector bundle theory again), but I cannot find a suitable $G$ so far. Therefore I am at a loss at how Browder can so easily see that this fact is true.
Any help would be appreciated!
 A: Here is a proof for when $X$ is a finite complex, modified from Milnor 'Microbundles Part I' Theorem 4.1:  
First observe that when $X$ is a sphere, this can be done: take a map $f$ from $X$ to itself of degree -1, an inverse of a fibration $\alpha$ is $f^* \alpha$.
Now for general $X$, assume without loss of generality that $X$ is connected and it has 1 0-cell only. Starting from the 0-cell, we shall construct the stable inverse to a spherical fibration $\alpha$ with fibre $\mathbb{S}^k$ inductively over the n-skeleta of $X$. Over the 0-cell the original spherical fibration $\alpha$ is automatically trivial itself. Assume an inverse $\xi_{n-1}$ is already defined on $X^{n-1}$, i.e. $\alpha|_{X^{n-1}} + \xi_{n-1}$ is trivial. Consider $\xi_{n-1} + \epsilon^k$ ($\epsilon^k$ is the trivial $\mathbb{S}^k$ fibration) on the boundary of each n-cell of $X$, it is fiber homotopy equivalent to $\alpha + \xi_{n-1}$ which is by assumption trivial, so the fibration can be extended over the n-cell. Doing this for all n-cells yields $\eta_n$ that is trivial over $X_{n-1}$. 
$\eta_n$ descends to a fibration on $X_n/X_{n-1}$, which is a wedge of spheres. By our special case, $\eta_n + f^* \eta_n$ is trivial for some $f$, pull it back to $X_n$ to get our $\xi_n$. This completed our induction. 
This obviously also proves it for the case when $X$ is homotopy equivalent to a finite complex. Is there a way to relax this assumption?
