Let $M$ be a connected closed manifold of dimension $n\geq 2$.


Can it happen that for any $I\subsetneq I\_n=\{1, \dots, n\}$ there are continuous maps $f, g:M\to M$ such that $f^\*=g^\*|\_{\oplus\_{i\in I}H^i(M, \mathbb{Z})}$ yet $f^\*\neq g^\*|\_{\oplus\_{i\in I\_n}H^i(M, \mathbb{Z})}$?



Manifold for which you need to specify the action on cohomology in each degree
I believe this cannot happen: trivially, if $\text{H}^1(M;\mathbb Z) = 0$, you clearly won't find a example of $f, g$ as desired with $I = \{2, \dots, n\}$.


Suppose first that $M$ is orientable. Pick $0 \neq \alpha \in \text{H}^1(M)$ that is not a proper integral multiple of some other cohomology class. Then, by Poincaré duality, we can find $\beta \in \text{H}^{n-1}(M;\mathbb Z)$ such that $\alpha \beta = \mu\_M$, the dual of the fundamental class $[M]$. Hence if $f, g \colon M \to M$ are such that $f\_{\ast} = g\_{\ast}$ in degrees below $n$, we also get $f\_{\ast} \mu\_M = (f\_{\ast} \alpha)(f\_{\ast} \beta) = (g\_{\ast} \alpha)(g\_{\ast} \beta) = g\_{\ast} \mu\_M$ and thus $f\_{\ast} = g\_{\ast}$ in all degrees; so for $I = \{1, \dots, n-1\}$ no $f,g$ as desired exist.


If $M$ is non-orientable, the same argument with $\mathbb F\_2$ coefficients should work.


