Is the intersection of boundaries of convex bodies a topological sphere? Let $K_1, K_2, \ldots K_n$ be convex bodies in $R^d$. Assume that for any index set $I$, $\cap_{i \in I} K_i$ is not empty and is not properly contained in any body $K_i$ for $i\in I$.
Is it true that $\cap_{i \in I} \partial K_i$ is a disjoint union of topological (or even better PL) spheres?
 A: A counterexample is as follows:
Let $K_1=\{x,y,z,w|x^2+y^2\leq 1\}$ and $K_2=\{x,y,z,w|z^2+w^2\leq 1\}$. The boundaries are just what you get when you replace the inequality with an equality, so their intersection is a torus, $x^2+y^2=1$, $z^2+w^2=1$.
Then the interiors intersect, for instance at the origin, fulfilling Patricia Hersh's condition. The boundaries are topologically transverse, fulfilling Richard Kent's condition. I cannot think of any additional reasonable condition that would disallow this counterexample.
A: Here's a way of producing any compact subset $X$ of $S^n$ as the intersection of the boundaries of two convex bodies in $\mathbb R^{n+1}$. The first convex body is simply the unit ball $B^{n+1}$, with boundary $\partial B^{n+1}=S^n$.
Te second convex body, call it $K$, is constructed as follows.
Consider a function $f:S^n\to \mathbb R$ such that $f^{-1}(0)=X$.
Such functions exist in great abundance, see e.g. the answers to this question.
By carefully selecting $f$ (i.e. by taking it to be small, and with small first and second derivatives), we can make sure that
$$
K:=\{x\in\mathbb R^{n+1}:\|x\|\le 1+f(x/\|x\|)\}
$$
is convex.
It is then clear, by construction, that $\partial B^{n+1}\cap \partial K = X$.

If $f$ admits both positive and negative values (which can be arranged iff the complement of $X$ in $S^n$ is disconnected), then neither of $B^{n+1}$ or $K$ is contained in the other one.
A: To get a "YES" answer, you have to assume that 
at any point of $p\in\partial K_i\cap \partial K_j$
any two supporting hyperplanes to $K_i$ and $K_j$ 
have angle $> \tfrac{\pi}2$.
The proof is by induction on $n$.
WLOG we may assume that all $\partial K_i$ are smooth.
Assume $S_{n-1}=\partial K_1\cap \partial K_2\cap \dots\cap K_{n-1}$ is a sphere.
Note that $f=\mathop{\rm dist}_{\partial K_n}$ is a concave function on $S_{n-1}\cap K_n$,
Perturb $f$ so it become smooth.
The function has one maximum point and by Morse Lemma the level set 
$S_n=f^{-1}(0)=\partial K_1\cap \partial K_2\cap \dots\cap K_{n}$ is a sphere.
A: Here is a well known counterexample.
Edit: Oh, sorry, a disjoint union of spheres is allowed. I misread the question.
