Order relation between cohomological dimensions of open orientable manifolds Let $M$ be an open orientable connected manifold and let $\operatorname{Cohdim}_{\mathbb{Z}_{2}}(M)$ and $\operatorname{Cohdim}_{\mathbb{Z}}(M)$ be the cohomological dimensions of $M$ over $\mathbb{Z}_{2}$ and $\mathbb{Z}$ respectively, where cohomological dimension is defined to be the largest degree of a non-trivial cohomology group with given coefficient ring.

Is there an order relation between $\operatorname{Cohdim}_{\mathbb{Z}_2}(M)$ and $\operatorname{Cohdim}_{\mathbb{Z}}(M)$? Is it true that $\operatorname{Cohdim}_{\mathbb{Z}_{2}}(M)\leq \operatorname{Cohdim}_{\mathbb{Z}}(M)$?

I know that this is true for closed orientable manifolds. Also, for closed non-orientable manifolds this is not true.
 A: It seems that your definition of cohomological dimension is not the standard definition, see here for example. The answer below is regarding cohomological dimension according to your definition.
First, you claimed that $\operatorname{Cohdim}_{\mathbb{Z}_2}(M) \leq \operatorname{Cohdim}_{\mathbb{Z}}(M)$ is not true for closed non-orientable manifolds $M$. That is incorrect. For such connected manifolds we have $H^{\dim M}(M; \mathbb{Z}_2) \cong \mathbb{Z}_2$ and $H^{\dim M}(M; \mathbb{Z}_2) \cong \mathbb{Z}_2$, so $\operatorname{Cohdim}_{\mathbb{Z}_2}(M) = \operatorname{Cohdim}_{\mathbb{Z}}(M) = \dim M$.
Moving on to the case of open manifolds, let $X$ be a topological space. From the short exact sequence of groups $0 \to \mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \to \mathbb{Z}_2 \to 0$, we obtain a long exact sequence in cohomology
$$\dots \to H^{k-1}(X; \mathbb{Z}_2) \to H^k(X; \mathbb{Z}_2) \xrightarrow{\times 2} H^k(X; \mathbb{Z}) \to H^k(X; \mathbb{Z}_2) \to H^{k+1}(X; \mathbb{Z}) \to \dots$$
Suppose that $\operatorname{Cohdim}_{\mathbb{Z}_2}(X) = k$. As $H^k(X; \mathbb{Z}_2) \neq 0$, it follows from the exactness of the above sequence that at least one of $H^k(X;\mathbb{Z})$ and $H^{k+1}(X; \mathbb{Z})$ is non-zero, so $\operatorname{Cohdim}_{\mathbb{Z}}(X) \geq k$ and hence
$$\operatorname{Cohdim}_{\mathbb{Z}_2}(X) \leq \operatorname{Cohdim}_{\mathbb{Z}}(X).$$
In general, the two are not equal, even for manifolds. In fact, the difference between the two can be arbitrarily large.
Example: Consider a $(2n+1)$-dimensional Lens space $L(p; q_1, \dots, q_{n+1})$ with $p > 1$ odd. Fix $x \in L(p; q_1, \dots, q_{n+1})$ and set $M = L(p; q_1, \dots, q_{n+1})\setminus\{x\}$. It follows from Mayer-Vietoris that
$$H^k(M; \mathbb{Z}) \cong \begin{cases}
\mathbb{Z} & k = 0\\
\mathbb{Z}_p & k \in \{2, 4, \dots, 2n\}\\
0 & \text{otherwise}.
\end{cases}$$
Note that for $k > 0$, the map $H^k(M; \mathbb{Z}) \xrightarrow{\times 2} H^k(M; \mathbb{Z})$ is an isomorphism because either $H^k(M; \mathbb{Z}) = 0$ or $H^k(M; \mathbb{Z}) \cong \mathbb{Z}_p$ and $\mathbb{Z}_p \xrightarrow{\times 2} \mathbb{Z}_p$ is an isomorphism since $(2, p) = 1$. It then follows from the long exact sequence above that
$$H^k(M; \mathbb{Z}_2) = \begin{cases}
\mathbb{Z}_2 & k = 0\\
0 & k > 0.
\end{cases}$$
So we have $\operatorname{Cohdim}_{\mathbb{Z}_2}(M) = 0$ and $\operatorname{Cohdim}_{\mathbb{Z}}(M) = 2n$.
