During some project, I stumbled over the following question: I know that the Euler-characteristic cannot be used to classify $3$-dimensional manifolds, as opposed to the $2$-dimensional case. However, one quite strong statement is that whenever we have a (compact, oriented) $3$-dimensional topological manifold $\mathcal{M}$ with boundary we have that
$$\chi(\mathcal{M})=\frac{1}{2}\chi(\partial\mathcal{M}).$$
As an example, if we take a $3$-manifold with boundary given by a $2$-sphere, we know that $\chi(\mathcal{M})=1$.
Now, during a research project, I have a topology given and I know the following properties: It is a compact $3$-manifold, its boundary is a $2$-sphere and it is orientable. Furthermore, I know that it has Euler characteristic equal to one. As already stated above, this does not necessarily imply that my topology is a $3$-ball. However, while searching in the internet, I was not able to find any other manifold having the properties written above. Is it possible to conclude that my topology is a $3$-ball, or does anyone know other examples with the given properties?
(For context: I am working with "crystallization theory". There you represent a simplicial complex by a proper 4-edge-coloured graph with some extra properties, which can be understood as the dual 1-skeleton of the complex. From such a graph, it is easy to read off the Euler-characteristic, because the (3-k)-simplices are in one to one correspondence with the "k-bubbles" of the graph, i.e. with the connected components of certain colours. Furthermore, there is a general theorem that states that a complex is orientable iff the graph is bibpartite. In general, the complex represented by such a graph is a pseudomanifold, but one can show that a graph gives rise to a PL-manifold, if and only if all connected components with 3 colours represent spheres or balls. Last but not least, one can construct a "boundary graph", which represents the dual 1-skeleton of the boundary complex. So, this is the reason why I know that my topology has all these properties.)
Best Answer
In order to conclude the discussion: Take any connected, orientable, closed (compact and without boundary) 3-manifold $X$ which is not $S^3$ and remove from it an open regular ball. The result is a compact connected, orientable 3-manifold $M$ with $\partial M=S^2$, $\chi(M)=1$, but $M$ is not homeomorphic (not even homotopy-equivalent) to the 3-ball. For instance, you can take $X$ equal the product of a closed connected orientable surface with $S^1$, or the real-projective 3-space, etc.