I was going through exercises in Hatcher to prepare for quals and then came across the following: (Ex 2.2.35)
Use Mayer-Vietoris to show that a finite simplicial complex $X$ for which $H_1(X)$ contains torsion cannot be embedded as a subspace of $R^3$ in such a way as to have a neighborhood homeomorphic to to the mapping cylinder of some map from a closed orientable surface to $X$.
I know how to prove the generalization (Corollary 3.46 in Hatcher)
If $X \subset R^n$ is compact and locally contractible then $H_i(X;Z)$ is $0$ for $i \geq n$ and torsion free for $i = n - 1$ and $i = n-2$
using Alexander duality and cohomology, but the exercise appears in Chapter 2 in Hatcher, before cohomology has even been developed!
Embarrassingly I've thought about this for hours but haven't got anywhere... I feel like I must be missing something basic. For the MV sequence I thought I could use $A$ and $B$ with $A \cup B = R^3$ and $A \cap B = X$ to get an isomorphism $H_1(X) \cong H_1(A) \oplus H_1(B)$ in reduced homology but it's still unclear how this could relate to a mapping cylinder, or why an open neighborhood would be relevant.
How does one go about showing the claim using only basic homological machinery (i.e. without appealing to results from cohomology)?
Assume that there's such and embedding of $X$ in $\mathbb{R}^3$ with a neighborhood $U$ homeomorphic to to the mapping cylinder of $f$ some map from a closed orientable surface $\Sigma$ to $X$.
Let $V = S^3 \setminus X$ and cover $S^3 = U \cup V$. Since $S^3$ has trivial homology in degrees 1 and 2, we conclude that $H_1(U\cap V) \cong H_1(U)\oplus H_1(V)$ from Mayer-Vietoris. Now, $X$ is a deformation retract of $U$, since $U$ is the mapping cylinder of $f$. Moreover, $U\cap V = \Sigma \times [0, 1)$. Then, since $H_1(U\cap V)$ is torsion-free, so is $H_1(U) \cong H_1(X)$.