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 $p\geq 3$?



geometric conditions on maps between manifolds inducing monomorphisms on cohomology
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.


