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Let $BM$ with one object, $*$ be the one-object category corresponding to the (unital, associative) monoid $M$. Is $BM \cong (BM)^{opp}$ as categories? If not, why?

I tried to prove that this is true but my professor states that this statement is not true. Here is my attempt:

I constructed the following functor, $F:BM \rightarrow (BM)^{opp}$ such that:

On Object: $F(*) = * \in$ Obj$((BM)^{opp})$

On Morphism: Let $f:* \rightarrow *$ be a morphism in $BM$. The functor acts on $f$ as $F(f) = f \in$ Mor$((BM)^{opp})$

On composition of morphisms: $F(f \circ g) = g \circ_{opp} f$

Now in the reverse direction, let $G:(BM)^{opp} \rightarrow BM$ be a functor such that:

On Object: $G(*) = * \in$ Obj$(BM)$

On Morphism: Let $f:* \rightarrow *$ be a morphism in $(BM)^{opp}$. The functor acts on $f$ as $G(f) = f \in$ Mor$(BM)$

On composition of morphisms: $G(f \circ_{opp} g) = g \circ f$

Trivially there exists a natural isomorphism $\epsilon: G \circ F \leftrightarrow \text{id}_{BM}$ and a natural transformation $\eta: \text{id}_{(BM)^{opp}} \leftrightarrow F \circ G$. This is becuase $(G \circ F)(f \circ g) = G(g \circ_{opp} f) = f \circ g$ and analogously for $F \circ G$.

So by defintion of equivalence categories, $BM$ and $(BM)^{opp}$ are equivalent categories.

Questions:

  1. Is this construction correct?
  2. Can contravariant functors like above $F, G$ be equivalence of categories? (or is it by definition, only covariant functors can be used in equivalence of categories?). Because my professor did not specify that contravariant functors cannot be used in the equivalence of categories.
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  • $\begingroup$ A contravariant functor $\mathcal{C} \to \mathcal{C}^\textrm{op}$ is a covariant functor $\mathcal{C} \to \mathcal{C}$. The contravariant functor you have constructed is just a disguised version of the identity functor, hence is not interesting. (When people say categories are equivalent or isomorphic without further specification, a covariant equivalence or isomorphism is meant.) $\endgroup$
    – Zhen Lin
    May 12, 2023 at 12:00

1 Answer 1

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Is this construction correct?

This does indeed construct a contravariant functor $(\mathbf{B}M)^{op} \to \mathbf{B}M$ and this functor is an equivalence, however...

Can contravariant functors like above $F,G$ be equivalence of categories? (or is it by definition, only covariant functors can be used in equivalence of categories?).

A contravariant functor $C \to D$ is a (covariant) functor $C^{op} \to D$ or equivalently $C \to D^{op}$. If such a functor is an equivalence, it means that $C^{op}$ is equivalent to $D$ or $C$ is equivalent to $D^{op}$.

If we did allow this to mean that $C$ is equivalent to $D$, then every category would be equivalent to its opposite, via the identity functor $C^{op}$ to $C^{op}$ (which is then a contravariant functor $C \to C^{op}$).

For this reason, it's uncommon to introduce a functor as a contravariant functor $C \to D$ and then say that it's an equivalence, since it isn't an equivalence between $C$ and $D$.

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