Homolgy of the 2d-skeleton of the 4d-cube 
I am currently trying to calculate the homology groups for the 2 dimensional skeleton of the 4 dimensional cube.
My problem is that I have no idea what this looks like; I know that the 1 skeleton is the hypercube graph, but that's really it. Once I have the skeleton, I'm guessing I can just use $\Delta$-complexes to figure out its homology.
Any help is appreciated! Thanks.
 A: Let $I = [0,1]$. Let $X$ be the 2-skeleton of $I^4$. First off $H_1 X = H_1 I^4 = 0$ (use long exact sequence of the pair $(I^4, X)$. Now consider the cellular chain complex of $X$ with coefficients in a field $F$. We have $0 \to C_2(X) \to C_1(X) \to C_0(X) \to 0$ which can easily be seen to be $0 \to F^{24} \to F^{32} \to F^{16} \to 0$. Now $H_0(X) = F$ as $X$ is connected. So $ker(F^{32} \to F^{16}) = F^{32-15}= F^{17}$. By our $H_1 X$ computation, we have $im(F^{24} \to F^{32})$  must be 17 dimensional. So we can conclude $H_2(X) = F^7$. 
Here's a tip on how to "see" what it looks like. The 0-cells of $I^4$ in a cubical decomposition are the points $x=(x_0,x_1,x_2,x_3)$ of $I^4$ where each $x_i$ is either $0$ or $1$. So there are $2^4$ of them. Each 1-cell is had by selecting a component to vary and the remaining components are either 0 or 1, e.g., the 2-nd component and a choice of $x_0,x_2,x_3$ in $\{0,1\}$, so $\{(x_0,t,x_2,x_3) \, : \, t \in [0,1]\}$. So we see that there are 4 ways to pick a component to have the parameter $t$, and there are $2^3$ ways to fill in the remaining components with 0's and 1's. Hence there are 32 1-cells in I^4. Similar combinatorics will show you what the 2-cells are. 
