Geometric intuition for torsion in $H_{2}$ of non-orientable $3$-manifold Let $M$ be a compact, connected $n$-manifold. Consider the homology groups $H_n(M)$ with coefficients in $\mathbb{Z}$.
It is well known that if $M$ is not $\mathbb{Z}$-orientable, then we have $H_n(M) =0$ and $H_{n-1}(M) = \mathbb{Z}/2 \oplus \mathbb{Z}^i$ for some $i \ge 0$.
The proofs are clear for me (remark: the main instruments used in the proof are the Universal Coefficients Theorem and the existence of an orientable double cover of $M$), but I'm quite curious if there exists a geometric/intuitive explanation for the torsion summand $ \mathbb{Z}/2$ of $H_{n-1}(M)$.
Can this phenomenon be visualized in the case of a non-orientable $3$-manifold or is this a purely algebraic result?
 A: Here is an extended version of my, now deleted, comment. 
Let $M$ be connected nonorientable a compact triangulated manifold (every topological 3-manifold admits a triangulation). A similar argument works for a manifold equipped with a CW complex structure, but it is less geometric in this case. I will work with simplicial homology. 
Let $c\in C_n(M; {\mathbb Z})$ denote the chain equal to the sum of all top-dimensional simplices. The boundary of this chain is a cycle $b=\partial c$ with even coefficients. Therefore, $a=\frac{1}{2}b$ is still a cycle with integer coefficients. Since $M$ is unorientable,   $a$ defines a nontrivial element of $H_{n-1}(M; {\mathbb Z})$. In order to prove this, you find a 1-cycle $e\in Z_1(M; {\mathbb Z}/2)$ (in the 1-skeleton of the dual triangulation) which has nonzero algebraic intersection number  with $a$: Take $e$ which reverses orientation. From this, you see that $[a]\ne 0$ in $H_{n-1}(M; {\mathbb Z})$ as well. On the other hand, clearly, $2[a]=[b]=0$. Hence, $[a]$ generates ${\mathbb Z}_2$ in $H_{n-1}(M; {\mathbb Z})$. 
I do not know how to see geometrically that $[a]$ generates a direct summand of  $H_{n-1}(M; {\mathbb Z})$. 
