I would like to use the normalized chain complex approach to compute the homology of $\Delta[1] / \partial \Delta[1] \times \Delta[1] / \partial \Delta[1]$.
In order to do that, I first have to understand what are the nondegenerate simplices of this product. I know that both terms have only one nondegenerate simplex in degree $0$ (the single vertex) and one nondegenerate simplex in degree $1$ (the single edge that is a loop on the vertex). However, when doing the product of simplices, nondegenerate simplices could come from degenerate simplices of $\Delta[1] / \partial \Delta[1]$.
Can you please help me understand what happens here, and what are the nondegenerate simplices? This is my first time doing an exercise of this kind, and I'm lost.
After that, I have to construct the normalized chain complex, so $$\cdots \to \mathbb{Z}[X^{\text{nd}}_n] \xrightarrow{d} \mathbb{Z}[X^{\text{nd}}_{n-1}] \to \cdots$$ where $d(x)$ is defined as the alternating sum of the $d_i(x)$, but in this case taking only the $d_i(x)$ that are nondegenerate.
I would really need some help understanding how to proceed on this kind of exercise. Thank you.
The $n$-simplices in the product of two simplicial sets $K$ and $L$ are of the form $(\sigma, \tau)$, where $\sigma$ and $\tau$ are $n$-simplices in $K$ and $L$, respectively. If $\sigma$ is actually $$ \sigma = s_I \overline{\sigma} := s_{i_1} s_{i_2} \dots s_{i_p} \overline{\sigma} $$ where $\overline{\sigma}$ is nondegenerate and $I=(i_1, \dots i_p)$, and similarly for $\tau = s_J \overline{\tau}$, then $(\sigma, \tau)$ will be nondegenerate if and only if $I$ and $J$ are disjoint. (This may not be obvious in this generality, but it is completely clear in the example you're asking about; more details there.)
In your case, to find (for example) the nondegenerate $2$-simplices in the product, we write down the $2$-simplices in each factor. If I let the left factor have vertex $v$ and $1$-simplex $e$, and if I let the right factor have vertex $w$ and $1$-simplex $f$, then the $2$-simplices in each are $$ s_0 s_0 v, \ s_0 e, \ s_1 e; \\ s_0 s_0 w, \ s_0 f, \ s_1 f. $$ If I want to form pairs of these, one from each factor, with disjoint degeneracies, I get two of them: $(s_0 e, s_1 f)$ and $(s_1 e, s_0 f)$. The other pairs are degenerate because the degeneracy maps are applied in each factor; for example, $(s_0 e, s_0 f) = s_0 (e, f)$, so this is degenerate.
Similarly you should get three nondegenerate $1$-simplices: $(s_0 v, f)$, $(e, s_0 w)$, and $(e,f)$.
All of this should match up with the simple "triangulation" of the standard model for the torus: take a rectangle with edges identified, and draw its diagonal to divide it into two triangles.
The face maps are also applied to each factor — $d_i (a,b) = (d_i a, d_i b)$ — so you should be able to compute the boundary maps in the associated chain complex from this.