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A basic result in the theory of model categories is that simplicial sets form a simplicial model category. The same is true for simplicial $k$-algebras. I have two questions related to this. One involves defining model categories via gluing and the second involves defining model categories via functor of points. I have no idea if these questions are extremely easy or difficult. I am sure they display my lack of knowledge when it comes to model categories.

Consider the category $\mathbb{V}_k$ of all separated schemes of finite type over a field $k$ along with a grothendieck topology $\mathcal{T}$ on $\mathbb{V}_k$.

Question A) Is the simplicial cateogry $s\mathbb{V}_k$ of $\mathbb{V}_k$ a simplicial model category (or even just a model category)?

Question B) Let $\mathbb{S}_k$ be the category formed by subfunctors of representable functors from $\mathbb{V}_k \to Sets$. Let $s\mathbb{S}_k$ be the simplicial category of $\mathbb{S}_k$. Is $s\mathbb{S}_k$ a simplicial model category?

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    $\begingroup$ Any category with finite (or small, to taste) limits and colimits is automatically a model category in a trivial way. You should at least specify what kind of weak equivalences you wish to consider. $\endgroup$
    – Zhen Lin
    Commented Sep 19, 2013 at 22:17
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    $\begingroup$ "Is $X$ a model category" is a type error. Being a model category is a structure, not a property. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 19, 2013 at 22:44
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    $\begingroup$ The weak equivalences should be the ones induced from the weak equivalences in the category of simplicial $k$-algebras. $\endgroup$
    – Andrew Stout
    Commented Sep 19, 2013 at 23:32
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    $\begingroup$ Regarding A you might have a look at Quillen's Homotopical Algebra, Section II.4. There are stated general conditions when simplicial objects form a simplicial model category. $\endgroup$
    – Lennart Meier
    Commented Sep 19, 2013 at 23:43
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    $\begingroup$ Prof. Meier: Thanks! I will try to prove that it satisfies these conditions. $\endgroup$
    – Andrew Stout
    Commented Sep 20, 2013 at 0:24

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