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Let $\cal C$ be a model category which is also additive. Suppose that the homotopy category $\operatorname{Ho}\mathcal C$ is additive, for example this is true when the weak equivalences in $\cal C$ is closed under biproducts (see this question).

If we take a cofibrant object $X$ and a fibrant object $Y$ then there is a natural isomorphism $$ \operatorname{Ho}\mathcal C(X,Y) \cong \mathcal C(X,Y)/\sim $$ where $\sim$ is the homotopy relation. Is this always a group isomorphism?

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2 Answers 2

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If by "additive" you mean an $\mathbf{Ab}$-enriched category with a zero object and biproducts, then yes. Let $\mathcal{M}$ be model category that is additive in this sense, let $\mathcal{M}_c$ be the full subcategory of cofibrant objects, let $\mathcal{M}_f$ be the full subcategory of fibrant objects, and let $\mathcal{M}_{cf} = \mathcal{M}_c \cap \mathcal{M}_f$. Here are the relevant facts: $\DeclareMathOperator{\Ho}{Ho}$

  1. The coproduct of a family of cofibrant objects is automatically a homotopy coproduct, so the localising functor $\mathcal{M}_c \to \Ho \mathcal{M}$ preserves coproducts. Dually, the localising functor $\mathcal{M}_f \to \Ho \mathcal{M}$ preserves products.
  2. Hence, the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ preserves the zero object and biproducts. (Note that $\mathcal{M}_{cf}$ is an additive subcategory of $\mathcal{M}$.)
  3. A category with a zero object and biproducts is automatically enriched over commutative monoids in a unique way, and a functor that preserves zero objects and biproducts is similarly enriched. Thus, there is a unique enrichment of $\Ho \mathcal{M}$ over commutative monoids that makes the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ an enriched functor.
  4. Since $\mathcal{M}_{cf}$ is actually $\mathbf{Ab}$-enriched and the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ is full and essentially surjective on objects, $\Ho \mathcal{M}$ is also $\mathbf{Ab}$-enriched.

Now, let $X$ and $Y$ be any two objects in $\mathcal{M}$. In order for the hom-set map $$\mathcal{M}(X, Y) \to \Ho \mathcal{M}(X, Y)$$ to be a group homomorphism, it is sufficient that the localising functor $\mathcal{M} \to \Ho \mathcal{M}$ preserve either the coproduct $X + X$ or the product $Y \times Y$. (We already know that it preserves initial and terminal objects.) Thus it suffices to take either $X$ cofibrant or $Y$ fibrant.

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Why did you start with "if by additive you mean..."? Is there any other definition of additive commonly in use? –  Omar Antolín-Camarena Nov 14 '13 at 3:54
    
Not that I'm aware of, but it's always good to be explicit! (By contrast the phrase ‘preadditive category’ seems to have at least two common definitions.) –  Zhen Lin Nov 14 '13 at 8:08

Yes, because the projection functor from the model category to the homotopy category preserves coproducts of cofibrant objects. That is actually the way of showing that the homotopy category has coproducts.

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