# n-skeleton product preserving.

It is true that the n-skeleta functor

$$sk_{n}:sSet\rightarrow sSet$$ is product preserving i.e., $sk_{n}(X\times Y)$ is naturally isomorphic to $sk_{n}(X)\times sk_{n}(Y)$.

• I think you need $sk_n(sk_n(X)\times sk_n(Y))$. Sep 16, 2018 at 20:17

No, this is false. For a very simple example, consider $$X=Y=\Delta^1$$ and $$n=1$$. Then $$sk_1(X)\times sk_1(Y)=\Delta^1\times \Delta^1$$ has a nondegenerate $$2$$-simplex, so it cannot be isomorphic to the $$1$$-skeleton of anything.
What is true is that there is a natural isomorphism between $$sk_n(X\times Y)$$ and $$sk_n(sk_n(X)\times sk_n(Y))$$. Probably there is some abstract nonsense way of seeing this with adjoint functors, but here is a very concrete explanation. We can identify $$sk_n(X)\times sk_n(Y)$$ with the simplicial subset of $$X\times Y$$ generated by pairs $$(a,b)$$ where $$a$$ is a simplex of $$X$$ which is a degeneracy of a simplex of dimension $$\leq n$$ and $$b$$ is a simplex of $$Y$$ which is a degeneracy of a simplex of dimension $$\leq n$$. So, $$sk_n(sk_n(X)\times sk_n(Y))$$ is the simplicial subset of $$X\times Y$$ generated by such pairs $$(a,b)$$ where additionally $$a$$ and $$b$$ themselves have dimension $$\leq n$$ (so that this pair is a simplex in the product of dimension $$\leq n$$). But such pairs $$(a,b)$$ are exactly all the simplices of $$X\times Y$$ of dimension $$\leq n$$, since every simplex of dimension $$\leq n$$ is a degeneracy of a simplex of dimension $$\leq n$$ (namely, itself). So $$sk_n(sk_n(X)\times sk_n(Y))=sk_n(X\times Y)$$ as simplicial subsets of $$X\times Y$$.