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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.

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    $\begingroup$ I think that you can define at least a “degree mod $2$“, as the mod $2$ cardinality of a general fiber of $f$. $\endgroup$
    – Aphelli
    Oct 17, 2020 at 12:37
  • $\begingroup$ 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. $\endgroup$
    – Sumanta
    Oct 17, 2020 at 13:00
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    $\begingroup$ You mean $f_*([M]) = \deg(f)[N]$. $\endgroup$
    – Paul Frost
    Oct 17, 2020 at 16:21
  • $\begingroup$ Yeah, I mean $f_*([M])=\text{deg}(f)[N]$. Good point I should edit to avoid confusion. $\endgroup$
    – Sumanta
    Oct 17, 2020 at 16:23

1 Answer 1

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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 $\deg(f)=\deg(\pi_1)$.

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