[Example 3.11, Hatcher]Cellular coboundary maps with $\mathbb{Z}$ coefficients are zero then so are those with arbitrary coefficients.

Question

In Example 3.11 of Hatcher's Algebraic Topology, it says that

All cellular boundary maps for $T^n$ with $\mathbb{Z}$ coefficients must be trivial, otherwise the cohomology groups would be smaller than computed above. Hence all cellular coboundary maps with arbitrary coefficients are zero, and the map $H^n (T^n, \dot{T^n} ; R) \rightarrow H^n (T^n ; R)$ is an isomorphism.

But I can't point out why cellular coboundary maps with $\mathbb{Z}$ coefficients are zero then so are those with arbitrary coefficients.

Answer 1

Either as a theorem or a definition you have that cellular cochain complex is naturally isomorphic to the cellular chain complex dualized by taking maps into some abelian group $G$. The boundary map in the dual cochain complex is just given by precomposing your function with the boundary map in your original chain complex. Precomposing with the zero map is always zero, so the boundary is zero.

Answer 2

If $f$ is the zero map between two abelian groups, then for any abelian group $G$, $\mathrm{Hom}(f, G)$ is also the zero map.