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