My question is not difficult, but I'm confused with something.

The question: Prove that if $F: \mathbb{C} \rightarrow \mathbb{D}$ is a equivalence of categories and $Z \in \mathbb{C}$ is a zero object then $F(Z)$ is a zero object.

I used the fact that $F$ is a equivalence iff $F$ is full, faithful and essentially surjective.

Given any other object $Y \in \mathbb{D}$ there exist some object $X \in \mathbb{C}$ such that $F(X) \simeq Y$. By the hypothesis, I have two unique morphisms $f: X \rightarrow Z$ and $g:Z \rightarrow X$, this imply that i have unique morphisms $F(f): F(X) \rightarrow F(Z)$ and $F(g): F(Z) \rightarrow F(X)$. If i denote $\sigma$ the isomorfism $F(X) \simeq Y$, then every morphism that i take in for exemple $Hom_{\mathbb{D}}(Y,F(Z))$ is in the form $F(f) \circ \sigma $.

If a change the isomorphism $\sigma$ (in case it is posible) for another isomorphism, then the conclusion is the same but with the other isomorphism and the result don't follow immediately, I'm missing something?

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    $\begingroup$ The conclusion is the same, regardless of which isomorphism you pick, namely, that $\hom_{\mathbb{D}}(Y,F(Z))=\{F(f)\circ \sigma\}$ and similarly for the map out of $F(Z)$. I don't see the problem: you would just say the same thing about a different isomorphism $\sigma'$. $\endgroup$ – Kevin Carlson Jun 19 '14 at 20:53
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    $\begingroup$ I'm dumb... ok $F(f)$ is constant in the isomorphisms..... is just the same guy whatever the isomorphism... $\endgroup$ – user158082 Jun 19 '14 at 20:57

You can also view this as a consequence of the following general facts (which are useful anyway):

  • Left adjoint functors preserve colimits
  • Dually: Right adjoint functors preserve limits
  • Equivalences are left adjoint and right adjoint.

It follows that every equivalence preserves initial objects and terminal objects, hence also zero objects.


The following proposition is quite useful.

Proposition. Let $\mathscr C\xrightarrow{F}\mathscr D$ be a functor. Then $F$ is an equivalence if and only if $F$ both of the following hold.

  1. For every $X,Y\in\DeclareMathOperator{Ob}{Ob}\Ob(\mathscr C)$ the map $\DeclareMathOperator{Hom}{Hom}\Hom_{\mathscr C}(X,Y)\xrightarrow{\Phi}\Hom_{\mathscr D}(F(X),F(Y))$ given by $\Phi(f)=F(f)$ is a bijection.

  2. For every $D\in\Ob{\mathscr D}$ there exists a $C\in\Ob(\mathscr C)$ such that $F(C)\simeq D$.

Now, suppose $\mathscr C\xrightarrow{F}\mathscr D$ is an equivalence and that $Z\in\Ob(\mathscr C)$ is a zero-object of $\mathscr C$. Let $D\in\Ob(\mathscr D)$ so property 2 of the proposition ensures an isomorphism $F(C)\xrightarrow{\phi}D$.

Property 1 of the proposition ensures a bijection $\Hom_{\mathscr C}(Z,C)\to\Hom_{\mathscr D}(F(Z),F(C))$. Moreover, the map $\Hom_{\mathscr D}(F(Z),F(C))\to\Hom_{\mathscr D}(F(Z),D)$ given by $f\mapsto \phi\circ f$ is a bijection (prove this!). This gives a bijection $\Hom_{\mathscr C}(Z,C)\to\Hom_{\mathscr D}(F(Z),D)$ so there exists exactly one $\mathscr D$-morphism $F(Z)\to D$.

A similar argument shows that there exists exactly one $\mathscr D$-morphism $D\to F(Z)$. Hence $F(Z)$ is a zero-object of $\mathscr D$.


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