# Degree of a map $\Bbb RP^n\to S^{n-1}\times S^1$

Let $$f:\Bbb RP^n\to S^{n-1}\times S^1$$ be a continuous map, where $$n\geq 2$$, and consider the induced map $$f_*:H_n(\Bbb RP^n)\to H_n(S^{n-1}\times S^1)$$ on homology. For even $$n$$, $$H_n(\Bbb RP^n)=0$$ so this map is trivial. For odd $$n$$, this map is $$\Bbb Z\to \Bbb Z$$. Can the degree of $$f$$, i.e. $$f_*(1)$$ can be an arbitrary integer? Or are there restrictions that $$\deg f$$ should satisfy?

Let $$n$$ be odd and $$n\ge 2$$.
We may consider the induced map $$f^\ast: H^\ast(S^{n-1}\times S^1)\to H^\ast(\Bbb RP^n)$$ on cohomology rings. If $$a$$ generates $$H^1(S^1\times S^{n-1})$$ and $$b$$ generates $$H^{n-1}(S^{n-1}\times S^1)$$, then their cup product $$a\smile b$$ generates the top cohomology, and we must have $$f^\ast(a\smile b)=f^\ast(a)\smile f^\ast(b)$$ But $$f^\ast(a)\in H^1(\Bbb RP^n)$$ and $$f^\ast(b)\in H^{n-1}(\Bbb RP^n)$$, and $$H^1(\Bbb RP^n)\cong 0$$. Therefore $$f^\ast(a)=0$$, so the induced map on the top cohomology must be trivial.
If $$f_\ast$$ were non-trivial, then it's a multiplication by $$k\neq 0$$. Universal coefficient theorem gives natural isomorphisms $$H^n(\Bbb RP^n)\cong \text{Hom}(H_n(\Bbb RP^n),\Bbb Z)$$ (and similarly for $$S^{n-1}\times S^1$$). Naturality allows us to conclude that $$f^\ast$$ must also be a multiplication by $$k$$, but we just saw that $$k=0$$ from the first paragraph, so our assumption that $$k\neq 0$$ must be wrong. So every continuous map induces trivial homomorphism on top homology.
Here's another way. Any such degree $$k$$ map gives a composition $$S^n\to \mathbb{R}P^n\to S^{n-1}\times S^1.$$ The first map has degree $$2$$ when $$n$$ is odd, so the composition has degree $$2k$$. Assuming $$n>1, \pi_n(S^1) = 0$$, so the composition above is homotopic to a composition $$S^n\to S^{n-1}\times\{1\}\hookrightarrow S^{n-1}\times S^1$$ which induces the zero map in the top homology because $$H_n(S^{n-1}) = 0$$. Therefore, $$2k = 0$$, hence $$k = 0$$. When $$n=1$$, we have maps of any degree.
Here's yet another way to see that the degree of a map $$f: \mathbb{R}P^n \to S^{n-1} \times S^1$$ is $$0$$ when $$n \geq 2$$.
For $$n \geq 2$$, we have $$\pi_1(\mathbb{R}P^n) \cong \mathbb{Z}_2$$ and $$\pi_1( S^{n-1} \times S^1) \cong \begin{cases} \mathbb{Z} &\text{if} \; n \geq 3 \\ \mathbb{Z}^2 &\text{if} \; n = 2 \end{cases}$$ Hence the induced map $$f_*: \pi_1(\mathbb{R}P^n) \to\pi_1( S^{n-1} \times S^1)$$ is zero. By covering space theory, this implies that $$f$$ lifts to the universal cover of $$S^{n-1} \times S^1$$. The universal cover is $$S^{n-1} \times \mathbb{R}$$ when $$n \geq 3$$, and $$\mathbb{R}^2$$ when $$n=2$$. In either case, the universal cover is a space with $$H_n = 0$$. We conclude that $$f$$ induces the zero map on $$H_n$$.