Let $R$ be the ring which is generated by homeomorphism classes $[M]$ of compact closed manifolds (of arbitrary dimension) subject to the relations that
$$[F]\cdot [B] = [E]$$
if there exists a fibre bundle $F \to E \to B$, and
$$[M] + [N] = [M \cup N]$$
if $M$ and $N$ are of the same dimension. Clearly, $[pt]$ behaves as a unit and we can write $[pt]=1$. Moreover, since $[F] \cdot [B] = [F \times B] = [B \times F] = [B] \cdot [F]$, we see that $R$ is a commutative ring.


It is clear that the Euler characteristic defines a homomorphism $\chi : R \to {\mathbb Z}$. What else can we say about the ring $R$ ? What can we say if everything is required to be oriented and/or smooth etc.? Is the ring $R$ finitely generated?


Example: Since $S^1$ is a double cover of itself, we get
$[S^0] \cdot [S^1] = [S^1]$, but $[S^0] = 2$ and hence $[S^1]=0$. In particular, the classes of all mapping tori of homeomorphisms vanish in $R$ since they are fiber bundles over $S^1$.



Ring of closed manifolds modulo fiber bundles
Consider the variation where we ask for smooth manifolds and smooth fiber bundles. Then I claim that $R$ is not finitely generated. 


The starting observation is that if $F \to E \xrightarrow{p} B$ is a smooth fiber bundle then 


$$0 \to \text{ker}(p) \to T(E) \to p^{\ast}(T(B)) \to 0$$


gives a splitting of the tangent bundle of $E$. There are cohomological obstructions to such splittings existing in general, which we can compute. The upshot is that if $E$ is a simply connected (so it has no nontrivial covers) closed smooth manifold whose tangent bundle has no nontrivial subbundles, then $[E]$ does not participate in any of the interesting relations defining $R$, and in particular cannot lie in the subring of $R$ generated by manifolds of dimension smaller than $\dim E$. Hence to show that $R$ is not finitely generated it suffices to write down a sequence of such $E$ of arbitrarily large dimension.


But this is standard: we can take the even-dimensional spheres $S^{2n}$. First, observe that because $S^{2n}$ is simply connected, it has no nontrivial covering spaces, and in addition every real vector bundle over $S^{2n}$ is orientable, hence has well-defined Euler classes (after picking an orientation). Second, the Euler class $e(T)$ of the tangent bundle is $2$ times a generator of $H^{2n}(S^{2n})$, and in particular does not vanish. Since the Euler class is multiplicative with respect to direct sum, if $T = T\_1 \oplus T\_2$ is a nontrivial splitting of the tangent bundle then $e(T) = e(T\_1) e(T\_2)$. But the cohomology groups that $e(T\_1)$ and $e(T\_2)$ live in both vanish for $S^{2n}$; contradiction. Hence the tangent bundle of $S^{2n}$ admits no nontrivial splittings, and so $S^{2n}$ is not the total space of any nontrivial smooth fiber bundle of closed smooth manifolds. 


(Maybe this argument can be rescued in the topological setting using tangent microbundles?)


(Strictly speaking this argument's not quite complete: we also need to show that there aren't any interesting bundles with total space the disjoint union of $S^{2n}$ with something else. But fiber bundle maps $p : E \to B$ are open, so the image of $S^{2n}$ under such a map is a connected component of the base, and we can restrict our attention to this connected component without loss of generality. Then $E$ breaks up, as a fiber bundle, as a disjoint union of $S^{2n}$ and whatever else, and we can restrict our attention to $S^{2n}$ again without loss of generality. In other words, in the defining relations we can assume that $E$ and $B$ are both connected without loss of generality.) 


