In Hatcher’s, he states that the following three cross products are equivalent:
$H^i(I^i,\partial I^i;R)\times H^j(I^j,\partial I^j;R)\to H^n(I^n,\partial I^n;R)$
$H^i(T^i,\dot{T} ^i;R)\times H^j(T^j,\dot{T} ^j ;R)\to H^n(T^n,\dot{T} ^n;R)$
$H^i(T^i;R)\times H^j(T^j ;R)\to H^n(T^n;R)$
Here $i+j=n,I=[0,1]$, $T^n$ is a n-Torus,the dots denote deletion of the top-dimensional cell. And he says the first two are evidently equivalent since we can think of the torus as a quotient of a cube. But pairs $(I^n,\partial I^n)$ and $(T^n, \dot{T}^n)$ are not identical. So I don’t know why they are equivalent?
Moreover, he says that all cellular coboundary maps for torus are zero, thus the maps $H^n(T^n,\dot{T}^n; R) \to H^n(T^n;R)$ are isomorphism. I also don’t understand why this follows? Hope someone could help. Thanks!
1. By excision, $H^i(I^i, \partial I^i)$ is $H^i( C^i, \partial C^i)$ where $C^i$ is a i-cell. And so is $H^i(T^i,\dot{T^i})$ by deformation retracting its one-point-deleted i-cell to its boundary. Therefore these two are the same.
2. Consider the cellular coboundary map, $H^n(T^n,\dot{T^i}) = H^n(X^n,X^{n-1})$ by deformation retracting $\dot{T^n}$'s one-point-deleted n-cell to its boundary. Since the cellular coboundary map is zero $H^n(X^n,X^{n-1}) = H^n(X)$ i.e. $H^n(T)$. Here we done.