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Proof.
We construct inductively by dimension .
If then is a point and .
Assume that , is already constructed.
Consider the simplex
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Each its -dimensional face is dual to a -cell of
.
Fix a sufficiently large number .
First we define .
Each -face of is contained in a unique
affine -space in . Furthermore, the adjoint
faces cut the polyhedron .
Thus we may identify with and, therefore,
with .
By the induction assumption we already have .
We define to be equal
to the union of these for all faces
of . By the induction hypothesis (and since
was large enough) the choices
over different faces agree.
Our next step is to extend to the complement of
. For each face of
consider its outer normal cone
(e.g. if is a facet
then is a ray). We define
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In other words, we span the region above the normal cone
of a -face by the translates of the manifold .
We set . By now we have defined
everywhere, but .
Consider a facet of ,
e.g. the one sitting in the hyperplane .
Since is large enough, is small
enough and the intersection
is close enough to
the zero set of . By the induction
hypothesis this zero set can be deformed to .
We define ,
using this deformation.
We repeat the same procedure for all other facets of
.
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