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Let $\mathbb{F}$ be a [[field]].
> [!proposition] Proposition. ([[a map of nonzero degree induces an embedding of cohomology rings]])
> Let $f:M \to N$ be a [[continuous|map]] between [[compact]], ($\mathbb{Z}$-) [[(homological) orientation of a manifold|oriented]] [[manifold|manifolds]] of dimension $d$.
>
If the [[degree of a map between manifolds|degree]] of $f$ is nonzero, then [[singular (co)chain map and homomorphism induced by a continuous map|induced morphism]] on [[singular cohomology|cohomology]] with $\mathbb{F}$-[[(co)homology with coefficients|coefficients]] $f^{*}:H^{*}(N; \mathbb{F}) \to H^{*}(M; \mathbb{F})$ is [[injection|injective]]. This is powerful, because [[cup product|we know]] $f^{*}$ is a [[ring homomorphism|homomorphism]] of ([[graded ring|graded]]) [[ring|rings]]. Thus, if we know how to compute [[cup product|cup products]] in $H^{*}(M; \mathbb{F})$, then we can do so in $H^{*}(N; \mathbb{F})$ as well.
^proposition
> [!proof]- Proof. ([[a map of nonzero degree induces an embedding of cohomology rings]])
> Suppose $\alpha \in \operatorname{ker }(f^{*}: H^{k}(N;\mathbb{F}) \to H^{k}(M; \mathbb{F}))$. If $\alpha \neq 0$, then [[the perfect Poincare pairing|because the pairing]] $\langle -,- \rangle : H^{k}(N; \mathbb{F}) \otimes H^{d-k}(N; \mathbb{F}) \to \mathbb{F}$
> is [[nondegenerate bilinear form|nondegenerate]] (we're working over a [[field]], so [[free module|freeness]] necessarily holds), there exists $\beta \in H^{d-k}(N; \mathbb{F})$ with $\langle \alpha, \beta \rangle \neq 0$, i.e., $(\alpha \smile \beta)[N]\neq0$; and since we are working over a [[field]] $\mathbb{F}$ we are free to scale $\beta$ so that $(\alpha \smile \beta)[N]=1$. So $\text{deg }f=\big(\alpha \smile \beta \big)(f_{*}[M])=f^{*}(\alpha \smile \beta))[M]=(\underbrace{ f^{*}\alpha }_{ =0 }\smile f^{*} \beta)([M])=0$where we have used that $f^{*}$ is a [[ring homomorphism]].
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
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> ```
> [!frontlink]
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> FROM outgoing([[]])
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> ```