Notion of degree of a map from an orientable manifold to non-orientable manifold

Before writing my question, I want to write something that I know.

Let $$M$$ and $$N$$ be two closed(compact, without boundary) connected topological manifolds of dimension $$n$$. Now, if both are $$\Bbb Z$$-orientable(though we simply write orientable), then we know that $$H_n(M;\Bbb Z)\simeq \Bbb Z\simeq H_n(N;\Bbb Z).$$ Let $$[M]\in H_n(M;\Bbb Z)$$ and $$[N]\in H_n(N;\Bbb Z)$$ be two generators. Now, for any continuous map $$f:M\to N$$ we have an induced map $$f_*:H_n(M;\Bbb Z)\to H_n(N;\Bbb Z)$$ i.e. we have an integer, called degree, written as $$\text{deg}(f)$$ such that $$f_*:[M]\longmapsto \text{deg}(f)\cdot[N].$$

Now, in the case $$N$$ is non-orientable, we have $$H_n(N;\Bbb Z)=0.$$ So, we can not define the notion of the degree in the above way. But, we have orientation $$2$$-cover. That is there is a connected closed orientable manifold $$\widetilde N$$ and a $$2$$-fold covering map $$\varphi:\widetilde N\to N$$. Now, if we can lift our map $$f$$ to a map $$\widetilde f:M\to \widetilde N$$ i.e. $$\varphi\circ \widetilde f=f$$, then we talk about degree of $$f$$ i.e. we can define $$\text{deg}(f):=2\cdot \text{deg}(\widetilde f)$$. Possibly this the most natural way. Another motivation for defining this way is that for any $$n$$-fold covering map $$p:X\to Y$$ between two finite CW-complexes we have $$n\cdot \chi(X)=\chi(Y)$$. Though, in general, there is no relation between Euler-characteristic and degree of a map.

But this type of lifting is not possible, this needs to satisfy $$\varphi_*\big(\pi_1(\widetilde N)\big)\supseteq f_*\big(\pi_1(M)\big).$$ This is the necessary and sufficient condition of lifting.

From here my question starts.

$$1.$$ Is there any particular type of maps for which the above type of lifting is possible?

$$2.$$ If $$1.$$ is not in general true, is there any notion of degree of a map from a closed oriented manifold to another closed but non-oriented manifold?

Thanks, in advance, Any help will be highly appreciated.

• I think that you can define at least a “degree mod $2$“, as the mod $2$ cardinality of a general fiber of $f$. Oct 17, 2020 at 12:37
• Thanks for your comments. I know the notion of degree mod $2$. Any manifold is $\Bbb Z_2$-orientable, i.e. $H_n(M;\Bbb Z_2)=\Bbb Z_2$ for any closed $n$-manifold, both orientable and non-orientable. So this notion is not quite helpful. We can distinguish two maps only if one of them has $\text{deg}_{\Bbb Z_2}=0$ and the other has $\text{deg}_{\Bbb Z_2}=1$. For example, we can find two maps from $\Bbb S^n\to \Bbb S^n$ having the same mod $2$-degree but are not homotopic, as there integral degrees are not same. Oct 17, 2020 at 13:00
• You mean $f_*([M]) = \deg(f)[N]$. Oct 17, 2020 at 16:21
• Yeah, I mean $f_*([M])=\text{deg}(f)[N]$. Good point I should edit to avoid confusion. Oct 17, 2020 at 16:23

If we insist that (1) the degree of a composition is the product of degrees, and (2) the degree of an $$n$$-sheeted connected covering space is $$n$$, then there is a unique way to define the degree of $$f:M\to N$$ when $$N$$ is non-orientable.
If $$f$$ lifts to a map $$\tilde{f}$$, conditions (1) and (2) imply the degree of $$f$$ must be given by your formula $$\deg(f)=2\deg(\tilde{f})$$. In the case $$f$$ does not lift, we can form the fiber product $$\tilde{M} := \tilde{N}\times_N M$$, which will be a closed orientable manifold. Let $$\pi_1:\tilde{M}\to \tilde{N}$$, $$\pi_2:\tilde{M}\to M$$ be projection onto the first and second factor, respectively. Then $$\varphi\circ\pi_1=\pi_2\circ f$$, and by condition (2) we have $$\deg(\pi_2)=\deg(\varphi)= 2$$, so condition (1) implies $$\deg(f)=\deg(\pi_1)$$.