# Rugged manifold

It is well known that any compact smooth $m$-manifold can be obtained from $m$-ball by gluing some points on the boundary.

Is it still true for topological manifold?

• To proof the smooth case, fix a Riemannian metric and consider exponential map up to cutlocus.

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If your manifold has a topological handle decomposition you'd get what you want by finding a maximal tree in the dual 1-skeleton. My understanding is topological handles were worked out by Kirby and Siebenmann but I've never really understood the details or even the precise statements of the theorems. But that's where I'd start looking. –  Ryan Budney Oct 15 '10 at 0:49
To add on what Ryan said: Kirby-Siebenmann proved that every closed topological manifold of dimension > 4 has a CW-structure, so one can get a manifold by identifying boundary points of finitely many disks. Whether one disk suffices, I do not know –  Igor Belegradek Oct 15 '10 at 1:15

The answer is yes -- Morton Brown's mapping theorem says that for every closed (connected) topological $n$-manifold $M$ there is a continuous map $f$ from the $n$-cube $I^n$ onto $M$ which is injective on the interior of the cube (for manifolds with boundary, see Remark 1 below). This was proved in early sixties and can be found in M.Brown,"A mapping theorem for untriangulated manifolds," Topology of 3-manifolds, pp.92-94, M.K.Fort, Jr.(Editor), Prentice-Hall, Englewood Cliffs, N.J.,1962. MR0158374

Main idea of the proof is simple: "Use local PL structures to expand a small $n$-cell in $M$ gradually, until it becomes the whole manifold."

This can be realized the "infinite composition" of engulfings of finitely many points at a time. This argument is not too difficult, so I will try to sketch it.

First consider a closed $n$-cell $C$ in $M$ and a finite set $X=\{x_1,\ldots,x_k\}$ of points of $M$ disjoint from $C$ which we want to engulf. Assume that each $x_i$ lies in some open $n$-cell $U_i$ which intersects $C$. For each $i$, fix a PL structure on $U_i$, and join $x_i$ and some point $y_i\in \partial C$ by a PL arc $\alpha_i\subset U_i$ relative to this structure. If we assume $\dim M\geq 3$ (this is not a restriction because the theorem is clear for $\dim M=1$, and easily follows from the surface classification for $\dim M=2$), we may require that $\alpha_i$'s should be disjoint. The regular neighborhood $Q_i$ of $\alpha_i$ within $U_i$ is a closed $n$-cell, and $Q_i$'s can be made disjoint. Let $h$ be a homeomorphism of $M$ pushing $y_i$ towards $x_i$ within $Q_i$ for each $i$ which is identity outside $Q_i$'s. Then $X\subset h(C)$, and we can apply this process again to $h(C)$ and another finite set $X'\subset M\setminus h(C)$. Repeating this process we obtain a sequence of engulfing homeomorphisms $h_1, h_2,\ldots$.

We can arrage that the uniform limit $f\colon M\to M$ of the composition of engulfing homeomorphisms $f_n=h_n\circ\cdots\circ h_1$ exists and $f(C)=M$. If we choose sufficiently small $Q_i$'s in each stage, one can make sure that $f$ is injective on the interior of $C$ (indeed, we can arrange that for each interior point $x$ there is $n$ such that $f_n(x)=f(x)$).

Remark 1: In Brown's paper, one can find a corollary that "If $M$ is a compact connected manifold with nonempty boundary $B$, then there is a surjection $f\colon B\times [0,1]\to M$ that restricts to the identity on $B\times 0$ and is injective on $B\times[0,1)$". Notice also that the above theorem can be applied to the closed manifold $B$. Then it follows that any compact topological $n$-manifold (possibly with boundary) can be obtained by identifying some points in the boundary of the $n$-ball.

Remark 2: Berlanga's theorem extended the Brown's theorem to noncompact manifolds: This theorem states that for every (connected) $\sigma$-compact $n$-manifold $M$, there is a nice kind of surjection $I^n\to \overline{M}$ similar to Brown's map, where $\overline{M}$ denotes the end compactification of $M$.

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To expand on Ryan's comment: A handle decomposition of a manifold is a restricted type of quotient of a disjoint union of handles. An $n$-dimensional $k$-handle is by definition the manifold $B^k \times B^{n-k}$, with the decoration that $(\partial B^k) \times B^{n-k}$ is its lower boundary and $B^k \times (\partial B^{n-k})$ is its upper boundary. The allowed quotient is obtained by induction. The $k$-skeleton $M_k$ is the union of all of the $\le k$-handles. To make $M_{k+1}$, you identify part of $\partial M_k$ with all of the lower boundary of the $k+1$-handles using a closed embedding of the latter into the former. That is a complete definition, although maybe more conditions can also be added.
In addition to upper and lower boundary, the handles have upper and lower 1-dimensional radial foliations. The upper radial foliation is singular at the core of the handle and the lower radial foliation is singular at the co-core. There is a lemma that a handle decomposition can be extended radially. The lemma allows you to convert a handle decomposition of $M$ into a special kind of atlas of charts of $M$. It also allows you to convert the handle decomposition into a CW complex using the cores of the handles, or a dual CW complex using the co-cores.
Moreover, you can always simplify a handle decomposition of a closed $n$-manifold so that there is only one $0$-handle and only one $n$-handle. This answers Burago's question if $M$ has a handle decomposition.
It is easy to make a handle decomposition of any smooth or PL manifold (in the latter case, say, any triangulated PL manifold) where everything as described holds. For the topological case, it was proven by Kirby and Siebenmann that every $n$-manifold with $n \ge 5$ has a handle decomposition. Also every $n$-manifold with $n \le 3$ has a unique PL and a unique smooth structure. It was proven by Freedman that a 4-manifold with no PL structure (equivalently, no smooth structure) does NOT have a handle decomposition. So, the answer to Burago's question is yes for all closed manifolds except possibly for non-smoothable 4-manifolds. I suspect that the answer is still yes for all 4-manifolds.