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edited Apr 13, 2017 at 12:58

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For general $C$ (I will drop $W$ from the notation) this is not true even when $D$ is an infinite discrete category.

In the answer to this similar question this similar question I described how to construct a sequence of relative categories $C_0, C_1, C_2, \ldots$ with objects $X_i, Y_i \in C_i$ such that $X = (X_0, X_1, \ldots)$ and $Y = (Y_0, Y_1, \ldots)$ are isomorphic as objects of $\prod_i \mathrm{Ho}(C_i)$ but not as objects of $\mathrm{Ho}(\prod_i C_i)$.

We take $C = \coprod_i C_i$ and similarly we have objects $X$ and $Y$ isomorphic in $\mathrm{Ho}(C)^\mathbb{N}$ but not in $\mathrm{Ho}(C^\mathbb{N})$. Of course $\mathrm{Ho}(C^\mathbb{N})$ is the homotopy category of $L(C^\mathbb{N})$ so $X$ and $Y$ are not equivalent in $\mathrm{Ho}(C^\mathbb{N})$. I'm not sure whether $\mathrm{Ho}(C)^\mathbb{N}$ is the homotopy category of $L(C)^\mathbb{N}$ (probably not), but at least the functor $L(C)^\mathbb{N} \to \mathrm{Ho}(C)^\mathbb{N}$ reflects equivalences and hence $X$ and $Y$ are still equivalent in $L(C)^\mathbb{N}$ so $L(C^\mathbb{N}) \to L(C)^\mathbb{N}$ is not an equivalence.

For general $C$ (I will drop $W$ from the notation) this is not true even when $D$ is an infinite discrete category.

In the answer to this similar question I described how to construct a sequence of relative categories $C_0, C_1, C_2, \ldots$ with objects $X_i, Y_i \in C_i$ such that $X = (X_0, X_1, \ldots)$ and $Y = (Y_0, Y_1, \ldots)$ are isomorphic as objects of $\prod_i \mathrm{Ho}(C_i)$ but not as objects of $\mathrm{Ho}(\prod_i C_i)$.

We take $C = \coprod_i C_i$ and similarly we have objects $X$ and $Y$ isomorphic in $\mathrm{Ho}(C)^\mathbb{N}$ but not in $\mathrm{Ho}(C^\mathbb{N})$. Of course $\mathrm{Ho}(C^\mathbb{N})$ is the homotopy category of $L(C^\mathbb{N})$ so $X$ and $Y$ are not equivalent in $\mathrm{Ho}(C^\mathbb{N})$. I'm not sure whether $\mathrm{Ho}(C)^\mathbb{N}$ is the homotopy category of $L(C)^\mathbb{N}$ (probably not), but at least the functor $L(C)^\mathbb{N} \to \mathrm{Ho}(C)^\mathbb{N}$ reflects equivalences and hence $X$ and $Y$ are still equivalent in $L(C)^\mathbb{N}$ so $L(C^\mathbb{N}) \to L(C)^\mathbb{N}$ is not an equivalence.

For general $C$ (I will drop $W$ from the notation) this is not true even when $D$ is an infinite discrete category.

In the answer to this similar question I described how to construct a sequence of relative categories $C_0, C_1, C_2, \ldots$ with objects $X_i, Y_i \in C_i$ such that $X = (X_0, X_1, \ldots)$ and $Y = (Y_0, Y_1, \ldots)$ are isomorphic as objects of $\prod_i \mathrm{Ho}(C_i)$ but not as objects of $\mathrm{Ho}(\prod_i C_i)$.

We take $C = \coprod_i C_i$ and similarly we have objects $X$ and $Y$ isomorphic in $\mathrm{Ho}(C)^\mathbb{N}$ but not in $\mathrm{Ho}(C^\mathbb{N})$. Of course $\mathrm{Ho}(C^\mathbb{N})$ is the homotopy category of $L(C^\mathbb{N})$ so $X$ and $Y$ are not equivalent in $\mathrm{Ho}(C^\mathbb{N})$. I'm not sure whether $\mathrm{Ho}(C)^\mathbb{N}$ is the homotopy category of $L(C)^\mathbb{N}$ (probably not), but at least the functor $L(C)^\mathbb{N} \to \mathrm{Ho}(C)^\mathbb{N}$ reflects equivalences and hence $X$ and $Y$ are still equivalent in $L(C)^\mathbb{N}$ so $L(C^\mathbb{N}) \to L(C)^\mathbb{N}$ is not an equivalence.

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created Jul 4, 2014 at 19:40

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For general $C$ (I will drop $W$ from the notation) this is not true even when $D$ is an infinite discrete category.

In the answer to this similar question I described how to construct a sequence of relative categories $C_0, C_1, C_2, \ldots$ with objects $X_i, Y_i \in C_i$ such that $X = (X_0, X_1, \ldots)$ and $Y = (Y_0, Y_1, \ldots)$ are isomorphic as objects of $\prod_i \mathrm{Ho}(C_i)$ but not as objects of $\mathrm{Ho}(\prod_i C_i)$.

We take $C = \coprod_i C_i$ and similarly we have objects $X$ and $Y$ isomorphic in $\mathrm{Ho}(C)^\mathbb{N}$ but not in $\mathrm{Ho}(C^\mathbb{N})$. Of course $\mathrm{Ho}(C^\mathbb{N})$ is the homotopy category of $L(C^\mathbb{N})$ so $X$ and $Y$ are not equivalent in $\mathrm{Ho}(C^\mathbb{N})$. I'm not sure whether $\mathrm{Ho}(C)^\mathbb{N}$ is the homotopy category of $L(C)^\mathbb{N}$ (probably not), but at least the functor $L(C)^\mathbb{N} \to \mathrm{Ho}(C)^\mathbb{N}$ reflects equivalences and hence $X$ and $Y$ are still equivalent in $L(C)^\mathbb{N}$ so $L(C^\mathbb{N}) \to L(C)^\mathbb{N}$ is not an equivalence.