Sufficient condition for $\mathbb{Z}$-orientability Let $X$ be a topological $n$-manifold. Let's define a R-orientation on $X$ as a choice of generators $\alpha_{x}\in H_{n}(X,X\setminus\lbrace x\rbrace;R)$ that is consistent.
Suppose that $X$ is $\mathbb{Z}/n\mathbb{Z}$-orientable, for some $n\neq 2$. How can I show that this implies that $X$ is $\mathbb{Z}$-orientable ?
Thanks 
 A: I'm going to assume that the clash of variables in your question was inadvertent and that you intended to write something like "Suppose that $X$ is $\mathbb{Z}/k\mathbb{Z}$-orientable, for some $k\neq 2$.
The universal coefficients theorem provides natural homomorphisms $H_n(X,X-A;\mathbb{Z}) \to H_n(X,X-A;\mathbb{Z}/k \mathbb{Z})$. When the domain is isomorphic to $\mathbb{Z}$, for example when $A$ is a point or the interior of a closed disc, this homomorphism is an isomorphic copy of the quotient homomorphism $\mathbb{Z} \mapsto \mathbb{Z}/k \mathbb{Z}$. Using that $k \ge 3$, it follows that the universal coefficients homomorphism takes the unique generating pair of $H_n(X,X-A;\mathbb{Z}) \approx \mathbb{Z}$ bijectively onto a generating pair for $H_n(X,X-A;\mathbb{Z}/k \mathbb{Z}) \approx \mathbb{Z}/k\mathbb{Z}$, and furthermore this bijection is natural. The concept of "naturality" here refers to naturality with respect to various commutative diagrams that arise in the definition of orientability.
Using naturality over and over, you can prove that there is a closed path based at $x$ that reverses the generators of $H_n(X,X-\{x\};\mathbb{Z})$ if and only if there is a closed path that reverses the generators of $H_n(X,X-\{x\};\mathbb{Z}/k \mathbb{Z})$. Thus, $X$ is orientable over $\mathbb{Z}$ if and only if it is orientable over $\mathbb{Z}/k\mathbb{Z}$.
