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Since it's true for any simplicial model category, and "very nice" is not a technical term, I think it's better to delete it so that future readers don't think an extra hypothesis is necessary.
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David White
David White
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Suppose that we have a very nice simplicial model category $M$. The simplicial enrichment will be denoted by $map_{M}$. Let $f:A\rightarrow B$ be a morphism in the category $M$ such that $A$ is cofibrant. Suppose that for any fibrant object $R$, the induced map $map_{M}(B,R)\rightarrow map_{M}(A,R)$ is a weak homotopy equivalence of simplicial sets. Do we have that $f$ is a weak equivalence ?

Suppose that we have a very nice simplicial model category $M$. The simplicial enrichment will be denoted by $map_{M}$. Let $f:A\rightarrow B$ be a morphism in the category $M$ such that $A$ is cofibrant. Suppose that for any fibrant object $R$, the induced map $map_{M}(B,R)\rightarrow map_{M}(A,R)$ is a weak homotopy equivalence of simplicial sets. Do we have that $f$ is a weak equivalence ?

Suppose that we have a simplicial model category $M$. The simplicial enrichment will be denoted by $map_{M}$. Let $f:A\rightarrow B$ be a morphism in the category $M$ such that $A$ is cofibrant. Suppose that for any fibrant object $R$, the induced map $map_{M}(B,R)\rightarrow map_{M}(A,R)$ is a weak homotopy equivalence of simplicial sets. Do we have that $f$ is a weak equivalence ?

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Let
Let
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detecting weak equivalences in a simplicial model category

Suppose that we have a very nice simplicial model category $M$. The simplicial enrichment will be denoted by $map_{M}$. Let $f:A\rightarrow B$ be a morphism in the category $M$ such that $A$ is cofibrant. Suppose that for any fibrant object $R$, the induced map $map_{M}(B,R)\rightarrow map_{M}(A,R)$ is a weak homotopy equivalence of simplicial sets. Do we have that $f$ is a weak equivalence ?