Let $S_\bullet ^-$ and $S_\bullet ^+$ be simplicial sets of dimension respectively $k_-$ and $k_+$. I'm trying to prove that $S_\bullet ^-\times S_\bullet ^+$ has not dimension $\leq k_-+k_+-1$.
Take two nondegenerate simplices $\sigma_-\in S_{k_-}$ and $\sigma_+\in S_{k_+}$, i.e. a map $\sigma_-\times \sigma _+:\Delta^{k_-}\times \Delta^{k_+} \to S_\bullet ^-\times S_\bullet ^+$. The only way I see to get a (possibly nondegenerate) simplex in $S_{k_-+k_+} ^-\times S_{k_-+k_+} ^+$ is to precompose $\sigma_-\times \sigma _+$ with a map $\iota:\Delta^{k_-+k_+}\to \Delta^{k_-}\times \Delta^{k_+}$, which is induced by an injective map $\iota:[k_-+k_+]\to [k_-]\times [k_+]$.
Given any two simplicial set $S_\bullet$, $T_\bullet$, note two things:
- if $\sigma\in S_n$ is nondegenerate and $d:S_n\to S_{n-1}$ is the image of an injective map $[n-1]\to[n]$, then $d(\sigma)$ is nondegenerate;
- a simplex $(\sigma, \tau)\in S_n\times T_n$ is nondegenerate iff either $\sigma$ or $\tau$ (including both) is nondegenerate.
Hence to get a nondegenerate $(\sigma_-\times \sigma_+)\circ \iota$, we would need $\iota$ to be injective in at least one component; however I doubt that such a $\iota$ exists. Do you think that my argument can be adjusted or is totally useless? Thanks for any suggestion.
One of $\sigma\in S_n$ or $\tau\in S_n$ being non-degenerate is sufficient but not necessary for $(\sigma,\tau)\in S_n\times T_n$ to be non-degenerate. For example, $((0,1,1),(0,0,1))\in\Delta^2\times\Delta^2$ is non-degenerate even though both $(0,1,1)$ and $(0,0,1)$ are degenerate.
Recall that $(\sigma,\tau)\in S_n\times T_n$ is degenerate if and only if for some $i$ it is in the image of $(\sigma',\tau')\in S_{n-1}\times T_{n-1}$ under $s_i\colon S_{n-1}\times T_{n-1}\to S_n\times T_n$ for some $i$. But $s_i\colon S_{n-1}\times T_{n-1}\to S_n\times T_n$ has to commute with $s_i\colon S_{n-1}\to S_n$, $s_i\colon T_{n-1}\to T_n$ and the projections $S_\bullet\times T_\bullet\to S_\bullet$ and $S_\bullet\times T_\bullet\to T_\bullet$. Thus $s_i((\sigma',\tau')=(s_i(\sigma'),s_i(\tau'))$.
Say a simplex is degenerate at $i$ if it is in the image of $s_i$. It follows from the above that $(\sigma,\tau)$ is degenerate at $i$ if and only if both $\sigma$ and $\tau$ are degenerate at $i$. Equivalently, $(\sigma,\tau)$ is non-degenerate if the union of the indices at which $\sigma$ and $\tau$ are non-degenerate comprises all indicies.
If $\sigma$ is a non-degenerate $k$-simplex, then repeatedly applying face and degeneracy maps to $\sigma$ yields an $n$-simplix non-degenerate at any desired subset of at most $k$ many indices among the $n$.
Thus, if the dimensions of $S_\bullet$ and $T_\bullet$ are $k_-$ and $k_+$ respectively, then for any $n$ and any choice of at most $k_-$ and at most $k_+$ many indices of an $n$-simplex, there is a pair of $n$-simplices for which the first is non-degenerate at the at most $k_-$ many indices and the second at the at most $k_+$ many indices. The corresponding simplex in the product is the non-degenerate if and only if the union of the indices of non-degeneracy are all the indices. It follows that a non-degenerate $n$-simplex exists if and only if $n\le k_-+k_+$.
Explicitly, if $\sigma_\pm\in S^\pm_{k_\pm}$ are non-degenerate, then $\underbrace{s_0\circ\cdots\circ s_0}_{k_+}\sigma_-\in S^-_{k_-+k_+}$ is degenerate at precisely the first $k_-$ indices, while $s_{n-1}\circ \cdots s_{k_+-1}\sigma_+\in S^+_{k_-+k_+}$ is degenerate at the precisely the last $k_+$ indices. It follows from that $(\underbrace{s_0\circ\cdots\circ s_0}_{k_+}\sigma_-,s_{n-1}\circ \cdots s_{k_+-1}\sigma_+)\in S^-_{k_-+k_+}\times S^+_{k_-+k_+}$ is non-degenerate.