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Let $M,N$ be manifolds whose dimensions may be different. Let $f: M\longrightarrow N$ be a smooth map. What geometric conditions on $f$ can we impose such that the induced homomorphism $$ f^*: H^*(N;\mathbb{Z}_2)\longrightarrow H^*(M;\mathbb{Z}_2) $$ is injective?

What geometric conditions on $f$ can we impose such that the induced homomorphism $$ f^*: H^*(N;\mathbb{Z}_p)\longrightarrow H^*(M;\mathbb{Z}_p) $$ is injective for a prime $p\geq 3$?

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    $\begingroup$ If there is a section of $f$, then those homomorphisms are split monomorphisms. This is also true if there is a section after replacing $M$ and $N$ be homotopic topological spaces. $\endgroup$ – Jason Starr Oct 24 '15 at 14:46
  • $\begingroup$ @JasonStarr I agree with you: one way to have $f^*$ injective, is to require it to admit a left inverse, which is the same as requiring $f$ to admit, up to homotopy, a right inverse, i.e., a section. To see it more geometrically, one may work with the graph $\Gamma_f$ of $f$ inside the cylinder $M\times N$: then if there exists a deformation of the cylinder which makes $\Gamma_f$ projecting onto the second factor, your $f^*$ is injective. $\endgroup$ – Giovanni Moreno Oct 28 '15 at 5:44
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Are you looking for things like the following?

Suppose $M$ and $N$ are of the same dimension connected, closed and $\mathbb{Z}/p\mathbb{Z}$ orientable. Suppose that the $(\mathbb{Z}/p\mathbb{Z})$ degree of $f$ is non-zero. Then the map $f^*:H^*(N)\rightarrow H^*(M)$ is injective. This basically follows from Poincaré duality. Given a class $\eta\in H^k(N)$, there exists a dual class $\beta\in H^{n-k}(N)$ such that $\eta\smile \beta$ is a generator of $H^n(N)$. By assumption $f^*(\eta\smile\beta)\not=0$, hence $f^*(\eta)\not=0$ and $f^*$ is injective. If $f$ is not surjective, the degree is zero and the map is not injective. The non-vanishishing of the degree can be translated into more topological terms. For example, if there is a regular value with a single point in the preimage, the map is of non-zero degree for all $\mathbb{Z}/p\mathbb{Z}$ if the manifolds are orientable (and $\mathbb{Z}/2\mathbb{Z}$ always).

If $M,N$ are closed and oriented with $\dim N>\dim M$ then the map is never injective as the fundamental class of $N$ is mapped to zero.

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    $\begingroup$ You could say a bit more by this argument. Say $N$ has dimension $n$. The same cup-product calculation says that if $f^*$ is injective on $H^n(N,Z/p)$, it is injective in all degrees. The non-vanishing of $f^*$ on $H^n(N,Z/p)$ could be phrased as the existence of an n-dimensional submanifold of $M$ mapping with non-zero Z/p degree onto $N$. This follows from the existence of a section, or more generally of a `multi-section' of degree relatively prime to $p$. $\endgroup$ – Danny Ruberman Oct 28 '15 at 13:59

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