I've got a homework problem this week about proving that any finite simplicial 3-complex with the property that each point in the complex is contained in an open set homeomorphic to an open set $U$ in $\mathbb{R}^3$ has Euler characteristic $0$.
What I'm wondering is about what it means for there to be a point in the complex that does is not contained in an open set $U$ as above. Could someone give an example for such a complex?
Consider just a single $3$-simplex $\Delta^3$. The interior of the $3$-simplex is homeomorphic to an open ball in $\mathbb{R}^3$, so such a $U$ exists for any point in the interior. But no such $U$ exists for a point on the boundary. This is nontrivial to prove, but intuitively, any open neighborhood of a point inside one of the boundary triangles will look like a closed half-space in $\mathbb{R}^3$, rather than an open set in $\mathbb{R}^3$.
I'm not sure exactly what your definition is of "simplicial 3-complex" is, but if it just means "3-dimensional simplicial complex", then even worse things can happen. For instance, your simplicial complex could be a disjoint union of a 3-simplex and a 1-simplex. Then small neighborhoods of points in the 1-simplex look like an interval in $\mathbb{R}$, not like open subsets of $\mathbb{R}^3$.
Roughly speaking, if such a $U$ exists for every point, then every point needs to be nicely "surrounded" by $3$-simplices that glue together to look like an open subset of $\mathbb{R}^3$. This means that if a point is in the interior of a 2-simplex, there must be exactly two 3-simplices which have that 2-simplex as a boundary face, so that you have two half-spaces gluing together to look like $\mathbb{R}^3$. For vertices or points in the interior of a 1-simplex, it is rather more complicated to understand what the $3$-simplices around the point can look like to get an open neighborhood homeomorphic to $\mathbb{R}^3$.