I was so proud of this solution about finding the homology groups of the cube $I\times I \times I$ but then a friend of mine make me feel uncomfortable about it. My "proof" is this:
The homology groups of the square are $H_0(I\times I)=\mathbb{Z} $, $H_1(I\times I)=0 $ and $H_2(I\times I)=\mathbb{Z} $. Since $I \times I $ is a deformation retract of $ I \times I\times I$ we have $H_0(I\times I\times I)=\mathbb{Z} $, $H_1(I\times I\times I)=0 $ and $H_2(I\times I\times I)=\mathbb{Z} $.
My friend said this:
Your proof may be wrong since you are calculating the homology group of the shell of the cube and the square is a deformation retract of the cube with but not of the shell of the cube.
Who's wrong?
Thank you all in advance
I'm not sure why you stopped. It's also true that $I \times I$ retracts onto $I$, and $I$ retracts onto $pt$. In other words, they are all trivial.
If you meant to do $\partial (I^3)$, then the homology is right, but your argument is wrong.