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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 $sk_{n}(X)\times sk_{n}(Y)$.

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

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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 $X\times Y$.

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