algebraic-topologydifferential-topologygeometric-topologymanifolds
Is there a general homeomorphism classification of contractible topological 4-manifolds (possibly with boundary or noncompact)?
In the compact case, any such manifold has a homology 3-sphere as its boundary (according to Wikipedia). Homology 3-spheres can be classified (via geometrization of 3-manifolds), so maybe there is hope for a classification of the interiors also.
Let $f: S\to S$ be a homeomorphism of a compact connected surface (possibly with boundary).
Theorem.
If $\chi(S)<0$ then the mapping torus $M=M_f$ of $f$ is a Seifert manifold if and only if the mapping class of $f$ is periodic, i.e., $f$ is isotopic to a periodic homeomorphism.
If $\chi(S)\ge 0$ then $M$ is Seifert unless $S$ is the torus and eigenvalues of the action of $f$ on $H_1(S, {\mathbb R})$ are not roots of unity.
Proof. I will prove only Part 1 since the proof is already way too long. Recall that the surface $S$ admits Thurston-Nielsen decomposition with respect to $f$, i.e., a finite collection (possibly empty) of simple loops $L_i\subset S$ which are pairwise disjoint, pairwise non-isotopic and not isotopic to the boundary of $S$ (if there is any), such that:
a. $f$ preserves the multiloop $L=\cup_i L_i$. In particular, there exists $N$ such that $f^N$ preserves each loop $L_i$ and each component $S_j$ of $S\setminus L$.
b. The homeomorphism $f_j^N= f^N|S_j$ is homotopic to a homeomorphism $g_j: S_j\to S_j$ which is either periodic (by taking larger $N$ we can assume that such $g_j$ is the identity) or is pseudo-Anosov.
A homeomorphism $f$ is called reducible if $L$ is nonempty.
A very nice proof of this fundamental theorem can be found for instance in the book by Casson and Bleiler "Homeomorphisms of surfaces after Nielsen and Thurston."
Since a manifold is Seifert if and only if it has finite cover which is Seifert, we can assume that $S$ is oriented and $f=f^N$ (replacing $f$ with $f^N$ amounts to passing to a finite cover of $M$).
We now define tori $T_i$ which are mapping tori of the restrictions $f|L_i$. Then, by construction, the manifold $M=M_f$ is the union of submanifolds with boundary $M_j=M_{f|S_j}$, which meet along the tori $T_i$. As it was observed in the other answer, if $f|S_j$ is pseudo-Anosov, then $M_j$ is hyperbolic and, hence, $M$ cannot be a Seifert manifold in this case. (For instance, this can be seen from the fact that $\pi_1(M_j)$ has trivial center, while $\pi_1$ of a Seifert manifold always has nontrivial center, after passing to a finite cover, unless this cover is the 3-sphere.)
It remains to analyze the case when each $f|S_j$ is isotopic to the identity. Let $A_i=\eta(L_i)$ denote a small annular neighborhood of $L_i$ in $S$. Let $S_j'$ denote the complement in $S_j$ to all the annuli $A_j$ which it meets. By isotopying $f$ further, we can assume that $f_j=f|S_j'$ is the identity while the restriction $f|A_i$ is an iterated Dehn twist $D^{n_i}$ along the loop $L_i$. Then the mapping torus of $f_j$ is the product $S_j'\times S^1$. If some $n_i$ equals zero, then $f|A_i=Id$ and we can eliminate the loop $L_i$ from $L$ without changing topology of $M$ and periodicity (or lack of thereof) of $f$.
Each annulus $A_i$ has two boundary circles which I will denote $A_i^+, A_i^-$; accordingly, each manifold $M_{f|A_i}\cong A_i\times S^1$ has two boundary tori $T_{i}^+, T_i^-$ which are the mapping tori of $f|A_i^\pm$. Each loop $A_i^\pm$ is a boundary loop of a unique component of $$ \bigcup_j S_j' $$ which I will denote $S_i^+$ and $S_i^-$ accordingly. (Note that it might happen that $S_i^+=S_i^-$, for instance, if $L$ is a single loop which does not separate $S$.) Accordingly, the (mapping) torus $T_{i}^\pm=M_{f|A_i^\pm}$ is a boundary component of the product manifold $M_i^\pm=M_{f|S_i^\pm}$, the mapping torus of the homeomorphism $f$ restricted to $S_i^+$ or $S_i^-$. The manifold $M_i\pm$ has canonical product structure (since $f$ restricts to the identity on $S_i^\pm$). Therefore, for each torus $T_{i}^+$ we obtain a canonical system of generators of the homology group: "Horizontal'' loop $a_{i}^+$ corresponding to the loop $A_i^+\subset S$ and the "vertical'' loop $b_i^+$ corresponding to the $S^1$-factor of the decomposition $M_i^+=S_i^+\times S^1$. The same applies to the torus $T_{i}^-$, where we switch all pluses to minuses.
Now, consider these loops $a_i^\pm, b_i^\pm$ in the product $M_{f|A_i}=T^2\times [0,1]$. The loops $a_i^+, a_i^-$ are, of course, isotopic to each other in this product manifold, since they correspond to isotopic curves $A_i^+, A_i^-$ on the surface $S$. I will denote by $[a_i]$ their common homology class in this product manifold.
However, this is not the case for for the loops $b^+_i, b^-_i$: From the definition of the Dehn twist we obtain that $$ [b_i^-]= [b_i^+]\pm n_i [a_i] $$ the plus or minus depend on which boundary circle of $A_i$ we marked with $+$ and which with $-$; the number $n_i$ here is the power of the Dehn twist that we use.
The bottom line is that when under the gluing $M_i^+$ and $M_i^-$ along $A_i\times S^1$, the fibers of the (Seifert) fibration of $M_i^+$ do not match (up to isotopy) fibers of the Seifert fibration of $M_i^-$.
It is important to note here that the manifolds $M_i^\pm$ here admit unique, up to isotopy, Seifert fibrations, namely, the ones coming from their product decompositions $S_i^\pm \times S^1$, since each component of the surface $S\setminus L$ has negative Euler characteristic. (Here we are using the fact that $S$ is not the torus: cutting torus along a loop results in the annulus $A$ and the product $A\times S^1$ admits infinitely many, up to isotopy, circle fibrations.) This uniqueness theorem should be in Hempel's book on 3-manifolds and in Orlik's book on Seifert manifolds. In particular, up to isotopy, we can talk about the Seifert fibration of the manifold $M_i^\pm$.
The product regions $A_i\times S^1$ of course, admit infinitely many Seifert fibrations; to remedy this, we adjoin each $A_i\times S^1$ to the product manifold $S_i^+\times S^1$. As the result, $M$ is obtained by gluing product manifolds $M_j\cong S_j\times S^1$ along their boundary tori in such a fashion that all the gluing maps do not preserve the (up to isotopy) fibers of the circle fibrations of the manifolds $M_j$.
Now, we can finish the proof of Part 1 of the theorem with the following lemma which, I remember seeing in the book by Jaco and Shalen "Seifert fibered spaces in 3-manifolds", Memoirs of Amer. Math. Soc. 220 (1979).
Lemma. Suppose that $M$ is a 3-dimensional manifold obtained by gluing oriented Seifert manifolds $M_j$ along their incompressible boundary tori, where each $M_j$ admits a unique Seifert fibration. Then $M$ is Seifert if and only if all gluing maps preserve (up to isotopy) fibers of the Seifert fibrations.
Proof. I will skip the proof of one direction of this lemma since it is not needed and assume that one of the gluing maps does not preserve circle fibers. Let $T\subset M$ denote the incompressible torus corresponding to this gluing. Suppose, that $M$ is Seifert fibered; by looking at its fundamental group, it is clear that the base of this fibration has to be of hyperbolic type (i.e., it is a hyperbolic orbifold). It is a standard fact of the theory of Seifert manifolds (I am sure, it is in Jaco and Shalen) that every incompressible torus in such a fibration is "vertical'', i.e., isotopic to a torus foliated by Seifert fibers. To see the algebraic side of this statement, consider the short exact sequence of fundamental groups induced by Seifert fibration: $$ 1\to {\mathbb Z}\to \pi_1(M)\to B=\pi_1(O)\to 1, $$ where $O$ is the base-orbifold of the fibration and the fundamental group of $O$ is understood in the orbifold sense. Since the torus $T$ is incompressible, its fundamental group yields a subgroup $Z^2$ of $\pi_1(M)$. Projection of this group to $B$ has to be abelian, and, by hyperbolicity assumption on $O$, infinite cyclic. Therefore, the intersection of $Z^2$ with the normal subgroup ${\mathbb Z}$ of $\pi_1(M)$ is a free (abelian) factor of $Z^2$. In particular, $T$ admits a foliation by circles where each fiber is homotopic to the generic fiber of the Seifert fibration of $M$. By working more, one promotes this to an isotopy of $T^2$.
Instead of isotopying the torus we can isotope Seifert fibration itself. Therefore, splitting $M$ along $T$ results in one or two Seifert manifolds and the gluing map preserves Seifert fibrations. Continuing inductively with respect to all the boundary tori of the manifolds $M_i$, we obtain that each $M_i$ admits a Seifert fibration and gluing maps preserve these fibrations. Since Seifert fibration of each $M_i$ was unique (up to isotopy), we are done. QED
Here's an alternate proof which doesn't use invariance of domain. It also gives a slightly stronger result.
Theorem:
Let $M^n$ be compact without boundary. Then there is no immersion $f:M\rightarrow \mathbb{R}^n$.
Proof: (sketch). Assume for a contradiction there is such an $f$. Since $M$ is compact, $f$ is a closed map, that is, it maps closed sets to closed sets. To see this, let $F\subseteq M$ be closed. Then it's compact because $M$ is, so $f(F)$ is compact, hence closed. (Here, we just use the fact that $f$ is continuous).
Further, $f$ is an open map. That it, it maps open sets to open sets. To see this, note that it is enough to map really small open sets to open sets because $f(\bigcup U_i) = \bigcup f(U_i)$.
It's a fact (a consequence of the implicit function theorem, if I recall) that every immersion locally looks like the natural inclusion $\mathbb{R}^k\rightarrow \mathbb{R}^n$ into the first $k$ coordinates. (This uses the boundarylessness of $M$ - if $p$ is on the boundary, this part doesn't work.)
Said more precisely, given our immersion $f$ and a point $p\in M$, there is a chart around $p$ and around $f(p)$ so that in these chart coordinates, $f$ looks like the inclusion.
Now, in our case $k = n$ and then the natural inclusion is a homeomorphism. This implies open sets in these special charts are mapped to open sets, so $f$ is open.
Putting this together, we've now seen that $f$ is an open and closed mapping. Now, what is $f(M)$? It must be compact because $M$ is, but it must also be both open and closed in $R^n$ because $M$ is open and closed in itself. This implies $f(M) = \mathbb{R}^n$ since that's the only nonempty clopen subset of $\mathbb{R}^n$, but $\mathbb{R}^n$ isn't compact, so we've reached a contradiction.
Best Answer
If you work in topological category and assume that your manifolds are compact then any two compact contractible 4-manifolds $W_1, W_2$ with homeomorphic boundaries $\partial W_1, \partial W_2$ are homeomorphic. This follows for instance from the main classification result of Richard Stong:
R. Stong, Simply-connected 4-manifolds with a given boundary. Topology Appl. 52 (1993), no. 2, 161–167.
Roughly speaking, Stong extended Freedman's classification theorem to the case of compact simply-connected 4-manifolds with boundary.
(The special case of contractible manifolds might have been known earlier, I am not sure.)
Thus, if you think that 3-dimensional integer homology spheres are "classified" then so are compact topological contractible 4-manifolds. (Personally I regard the classification problem of integer homology spheres which are hyperbolic 3-manifolds to be hopelessly complicated. But modulo this problem, yes, we "know" what 3-dimensional homology spheres are.) Of course, a classification up to diffeomorphism of smooth 4-manifolds with the given boundary is well beyond reach at this point.