# What is wrong with the “naive” proof of the Hauptvermutung?

The Hauptvermutung is the statement that any two PL structures on a topological space have a common refinement. It is false in general, but (I think) true for some low dimensional manifolds.

The special case I'll consider is when the topological space is a smooth 2-dimensional manifold. In this case, the Hauptvermutung (roughly) states that for any two embeddings of two graphs into the manifold (such that the resulting faces are homeomorphic to disks), there is a graph (with an embedding) that contains both graphs as a subgraph that preserves that PL structure (cue the hand-waving).

It's "obvious" that if you take the union of the two graphs (again, embedded into the manifold) and add all the necessary points (such as the points at intersections of edges of the graphs), that you'll get a refinement common to both graphs (bar some subtle cases, like when the two graphs don't intersect)

In this answer, it's stated that this proof is wrong for "fractal" reasons. Does this mean that the proof still goes through if we restrict to finite graphs? If not, why?

This question is motivated by a problem posed a year or two back in my Differential Geometry class. The problem required that I prove that the Euler Characteristic (on a smooth, compact surface), defined by the quantity $V-E+F$ for any embedding of a graph, is well defined. I used essentially the proof above to construct a common refinement of any two graphs that preserved the quantity $V-E+F$. Since that particular homework was never returned, I never found out if my proof was correct, so now I'm wondering if I missed something.

EDIT: Since this approach clearly doesn't work (thanks to Alex Degtyarev in the comments), I might ask if there's any other way of proving that the Euler characteristic for compact surfaces defined in this way is well-defined. I know you can just use the homology groups (which is the preferred approach), but since we didn't mention homology in my class, I suspect there's a more elementary method of proof.

• Imagine one graph with an edge like $y=0$ and the other, with one like $y=x\sin(1/x)$. The union just won't be a graph. – Alex Degtyarev Dec 9 '14 at 10:04
• There's my problem: I never considered that the two graphs might intersect an infinite number of times. Thanks for that. – SCappella Dec 9 '14 at 22:09
• I think there is no way to do this in the topological category. You're going to have to first prove that topological surfaces can be given PL or smooth structures. Once you have that, you can use PL or smooth triangulations and apply transversality to make an argument like you propose go through. – Andy Putman Dec 9 '14 at 23:30

As pointed out in comments, your version of the Hauptvermutung cannot be true because there is no reason why the intersection of two graphs should be nice in any way. This has nothing to do with $C^0$ vs $C^\infty$. Since any closed subset of $\mathbb{R}^n$ is the zero set of a smooth function, you can have a pair of smooth graphs where two edges intersect along a very bad closed set.
Let's stick to smooth triangulations. It means there is a simplicial complex $K$ linearly embedded in some $\mathbb{R}^N$ and a homeomorphism $h$ from $K$ to your surface $S$ whose restriction to each simplex $\sigma$ is a smooth embedding (ie. if P denotes the affine subspace spanned by $\sigma$ in $\mathbb{R}^N$, there is an open neighborhood $U$ of $\sigma$ in $P$ and a smooth embedding $h_\sigma$ from $U$ to $S$ such that $h$ and $h_\sigma$ coincide on $\sigma$).
Now Whitehead's theorem (which holds in any dimension) says that if $h : K \to S$ and $g : L \to S$ are two such smooth triangulations then one can perturb $h$ and $g$ and get sub-complexes $K' \subset K$ and $L' \subset L$ such that $h^{-1}\circ g$ is a linear isomorphism from $L'$ to $K'$. Again: you cannot hope to get rid of the perturbation here (but you can make it $C^1$-small).
If you want to prove invariance of Euler characteristic using this path then you now have to prove a purely PL fact: if $K$ is a simplicial complex and $K'$ a subcomplex then $\chi(K) = \chi(K')$. Clearly it's enough to consider the case where $K$ is a simplex (a triangle in your case). One very nice trick is to extend any edge of $K'$ until it hits the boundary of $K$. This way you get a common refinement $K''$ which is obtained from $K$ and $K'$ by only one kind of operation: bissection (cutting one simplex using an hyperplane). This operation obviously preserves $\chi$. This trick can be used to get a very easy proof of invariance of simplicial homology. See also Siebenmann's proof of Reidemeister-Singer or Giroux's proof of the open book theorem for contact 3-manifolds for further use of this trick.
Further reading: Lurie's exscellent lecture notes for a proof of Whitehead's theorem. Moise's book Geometric Topology in Dimensions 2 and 3 if you are interested in the $C^0$ case from a traditional point of view. Hatcher's great expository paper about smoothing in dimension 2.