I think if you take into account the cup product structure, you can directly see that the action on cohomology is trivial.

Doesn't it follow immediately from Liouville? $f(x) = 1/f(x^{-1})$ together with the fact that $f$ has no zero at $0$ shows that it is bounded.

This is for the first statement only, right? (I.e., not the one about surjective or injective maps only, which I wouldn't expect to be true)

What about $c_k = \frac{1}{(k+1)^2}$?

You can compute $x^{2^n}$ mod $f$ by starting with $x$, and then repeatedly squaring and reducing modulo $f$ in each step. Takes you $n$ steps, and the degree of none of the intermediate values gets larger than twice the degree of $f$.

Just a comment: since any such map has image in a finite-dimensional $\mathbb{Q}$-subvector space in $\mathbb{R}$, and any such subspace surjects onto $\mathbb{Q}$, we can postcompose to get a map to the rationals with no more pairs violating addivity. By multiplying with the common denominator, we can further reduce to a map $\mathbb{F}_p\rightarrow\mathbb{Z}$.

To be more precise: A proof through a chain of relations that $g=e$ can be used to construct a homeomorphism $M″(g)\simeq M'$ that is the identity on $\pi_1$ under the identifications with $G$, and having such a homeomorphism proves $g=e$.

So the problem $g=e$ can be reduced to existence of a homeomorphism of $M''(g)\simeq M'$ which lifts the identity on $G$ under the identification $\pi_1(M''(g)) \simeq G$, $\pi_1(M')\simeq G$. Does this address both of your concerns?

The logic is as follows: Given $g\in G$, you can define $M''(g)$ which due to its construction comes with an isomorphism $\pi_1\simeq G$. If $g=e$, then you can algorithmically find a chain of relations exhibiting $g=e$, and this gives an isotopy from the sphere they surgered on to the trivial sphere, which gives you a homeomorphism $M''(g)\simeq M'$ compatible with the respective isomorphisms of $\pi_1$ to $G$. If $g\neq e$, then they can't be homeomorphic, since they have different $\pi_2$.

Oh, also, the paragraph before seems to answer OPs question in the negative. They construct two different manifolds with the same $\pi_1$ such that an algorithm checking whether they are homeomorphic would solve the word problem in $\pi_1$.

arxiv.org/pdf/math/9707232.pdf this paper claims to prove that the homeomorphism problem for simply connected manifolds of dimension $\geq 5$ is decidable (Theorem 1).

YCor, as I understand it, "matric" is the adjective of "matrix", so a Toda bracket of matrices is a "matric Toda bracket". merriam-webster.com/dictionary/matric

I'm afraid I don't know a nice reference for matric Toda brackets, but you can certainly introduce them as usual smash Toda brackets between spectra of the form $\bigvee_{i,j} S^{d_{ij}}$, using multiplication maps that look like the usual multiplication rules for matrices. After my edit, all the degrees are consistent in the example above. Alternatively (but that's less consistent with the argument in the post), describe them as composition Toda brackets between spectra of the form $\bigvee_i S^{d_i}$.

Looking back at this 3-year old post, I'm not actually sure whether inhomogeneous Toda brackets make sense, but noticed that setting the lower right entry to 0 makes this into perfectly fine homogeneous matric Toda brackets, so I edited it.