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Show that an equivalence of categories sends products to products and coproducts to coproducts. That is, if $X_i$ are a family of objects in $\mathcal{C}$ with coproduct $X$ then $F(X)$ is the coproduct of $F(X_i)$ in $\mathcal{D}$, and similarly for products.

Hint on how to get started?

Let me first say that I'm confused what really needs to be shown. a) Manipulate $F(X_i)$ etc directly, and construct the "connecting morphism" (I think you understand what I mean by that) somehow. b) Start with a collection $(Y_i)$ in $\mathcal{D}$ and any object $D$ equipped with morphisms $g_i : Y_i \rightarrow D$, and show that there exists a unique morphism $\theta':F(X) \rightarrow D$ such that $g_i = \theta' F(\alpha_i)$ (question is, can I really assume that such a $D$ and $g_i$ exist?). Anyway, here's what I've tried (full of holes)

Let $D$ be an object in $\mathcal{D}$ equipped with morphisms $g_i : Y_i \rightarrow D$. Let $F:\mathcal{C} \rightarrow \mathcal{D}$, $G:\mathcal{D} \rightarrow \mathcal{C}$ and put $X_i=G(Y_i)$, $f_i=G(g_i)$, $C=G(D)$. Then $(C,f_i)$ is such that there exists a unique morphism $\theta:X \rightarrow C$ such that $f_i=\theta \alpha_i$, where $\alpha_i : X_i \rightarrow X$. Let $\varepsilon : FG \rightarrow \mathrm{id}_{\mathcal{D}}$ be a natural isomorphism. This means we have isomorphisms $\varepsilon_{Y_i}:F(X_i)=FG(Y_i) \rightarrow Y_i$ and $\varepsilon_D:F(C)=FG(D) \rightarrow D$ such that $\varepsilon_D FG(g_i)=g_i \varepsilon_{Y_i}$. LHS equals $\varepsilon_D F(f_i)=\varepsilon_D F(\theta \alpha_i)=\varepsilon_D F(\theta) F(\alpha_i)$.

But yeah, this doesn't achieve much. All my ideas are along these lines but I can't get anything useful out of it. Many thanks for advice on this.

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1 Answer 1

You have to show exactly what the exercise says: If $\{X_i \to X\}$ is a coproduct diagram in $\mathcal{C}$, then $\{F(X_i) \to F(X)\}$ is a coproduct diagram in $\mathcal{D}$. There at least four proofs (of course related with each other) for this observation. I will sketch them.

  1. (Direct way) Let $G : \mathcal{D} \to \mathcal{C}$ be a functor with $F G \cong 1_\mathcal{D}$ and $GF \cong 1_{\mathcal{C}}$. Let $T \in \mathcal{D}$ and let $\{F(X_i) \to T\}$ be a family of morphisms. These induce morphisms $X_i \cong G(F(X_i)) \to G(T)$. Hence, there is a unique morphism $X \to G(T)$ such that the diagram ... commutes. This corresponds to a morphism $F(T) \to X$ such that the diagram ... commutes.

  2. (Using ess. surjectivity and fully faithfulness) We have to show that the canonical map $\hom(F(X),T) \to \prod_i \hom(F(X_i),T)$ is bijective, for all $T \in \mathcal{D}$. Since this is a natural transformation of functors in $T$ and $T \cong F(S)$ for some $S \in \mathcal{C}$, we may even assume $T=F(S)$. Since $F$ is fully faithful, the map identifies with $\hom(X,S) \to \prod_i \hom(X_i,S)$, which is bijective.

  3. (Adjunctions) More generally, if $F$ is left adjoint to some functor $G$, then $F$ preserves all coproducts (in fact all colimits). The reason is that $$\prod_i \hom(F(X_i),T) \cong \prod_i \hom(X_i,G(T)) \cong \hom(X,G(T)) \cong \hom(F(X),T).$$

  4. (Principle of equivalence) Every notion, property, object etc. defined in the language of category theory is preserved by equivalences of categories. Actually this is the main idea behind the definition of an equivalence of categories.

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(4) requires care: one must avoid talking about equalities of objects for that to be true. –  Zhen Lin Sep 24 '13 at 9:03
    
For me, equality of objects doesn't belong to the language of category theory. Equality refers to set theory. –  Martin Brandenburg Sep 24 '13 at 9:08

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