Eqiuvalence (Category Theory)

Let $\mathcal{C},\mathcal{D}$ be categories.

We say $\mathcal{C}$ and $\mathcal{D}$ are equivalent, written

$$\mathcal{C}\simeq \mathcal{D},$$

if there exists a functor $F:\mathcal{C}\to \mathcal{D}$ that is:

• full
• faithful
• essentially surjective

Remarks

This is weaker than isomorphism of categories.

A common slogan is that category theory studies structure only up to equivalence, rather than up to isomorphism.

One can also define equivalence by asking for functors

$$F:\mathcal{C}\to \mathcal{D},G:\mathcal{D}\to \mathcal{C}$$

together with natural isomorphisms

$$GF\cong 1_{\mathcal{C}},FG\cong 1_{\mathcal{D}}.$$

This is equivalent to the full, faithful, essentially surjective characterization above.

Some examples of equivalences use the Axiom of Choice. In particular, identifying each set with a chosen ordinal of the same cardinality depends on choice.

Example

Let Set be the category of sets and functions, and let Ord be the full subcategory on ordinals, with arbitrary functions as morphisms.

Assuming the Axiom of Choice, there is an equivalence

$$\mathbf{Set}\simeq \mathbf{Ord}.$$

More precisely, one chooses for each set $X$ an ordinal $\alpha (X)$ together with a bijection

$${\varphi }_X:X\stackrel{\sim }{\to }\alpha (X).$$

Define a functor

$$F:\mathbf{Set}\to \mathbf{Ord}$$

by

$$F(X)=\alpha (X),F(f)={\varphi }_Y\circ f\circ {\varphi }_X^{-1}.$$

Full

Given a function

$$h:\alpha (X)\to \alpha (Y),$$

define

$$g={\varphi }_Y^{-1}\circ h\circ {\varphi }_X.$$

Then

$$F(g)=h.$$

Faithful

Suppose $f,g:X\to Y$ satisfy $F(f)=F(g)$. Then

$${\varphi }_Y\circ f\circ {\varphi }_X^{-1}={\varphi }_Y\circ g\circ {\varphi }_X^{-1}.$$

Composing on the left with ${\varphi }_Y^{-1}$ and on the right with ${\varphi }_X$ gives

$$f=g.$$

Essentially Surjective

Let $\beta$ be an object of $\mathbf{Ord}$. Since $\beta$ is itself a set, $F(\beta )=\alpha (\beta )$, and by construction there is a bijection

$$\alpha (\beta )\cong \beta .$$

Hence every object of $\mathbf{Ord}$ is isomorphic to one in the image of $F$.

Resources

• https://math.stackexchange.com/questions/4476928/how-do-i-think-of-a-representable-functor