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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?

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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.

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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.

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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 $H_n$.

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