# Injective Representables

Posted on 7 April 2011 by Nicolas Wu
Tags: Category Theory

In this article I’ll show how the injective and surjective properties of a function have a nice relationship with the injectivity of the corresponding representable functors, and how the right cancellative definition of surjection can be derived from the standard one. This is a fairly trivial discussion about a simple observation, so don’t expect anything spectacular.

The definition of injectivity is usually given in the following terms, where a function is injective when it is left cancellative:

$$h$$ is injective iff $$\forall f, g \cdot h . f = h . g \Rightarrow f = g$$

Surjectivity can be described in similar terms, where a function is surjective when it is right cancellative:

$$h$$ is surjective iff $$\forall f, g \cdot f . h = g . h \Rightarrow f = g$$

While these two definitions show the similarity between the injective and surjective properties, this definition of surjectivity isn’t the standard one. The standard definition goes as follows:

$$h$$ is surjective iff $$\forall y \cdot \exists x \cdot h~ x = y$$

I don’t like this definition, since I don’t find it easy to work with in proofs, and it doesn’t show the relationship between surjections and injections well.

## Representables

The representables, or Hom-functors, are functions from arrows to arrows in a category. These functions come in two flavours, the covariant representable, and the contravariant representable. Given objects $$S$$ and $$T$$ in a category $$ℂ$$, we write $$ℂ(S, T)$$ denote the set of all arrows from $$S$$ to $$T$$.

### Covariant Representable

Given a function $$h : X \rightarrow Y$$, the covariant representable, $$ℂ(S,-)$$ is defined as:

ℂ(S,—) :: (X -> Y) -> ℂ(S, X) -> ℂ(S, Y)
ℂ(S,—) h f = h . f


As shorthand for the above, we usually slot the argument $$h$$ into the dash:

ℂ(S, h) = ℂ(S,—) h


### Contravariant Representable

Given a function $$h : X \rightarrow Y$$, the contravariant representable, $$ℂ(-,T)$$ is defined as:

ℂ(—,T) :: (X -> Y) -> ℂ(X, T) -> ℂ(Y, T)
ℂ(—,T) h f = f . h


Again, we use the following shorthand, where we slot $$h$$ into the dash:

ℂ(h, T) = ℂ(—,T) h


One feature of these representables that I like particularly is that they give rise to a clean correspondence between injectivity and surjectivity.

## Injectivity

The injective Representables give rise to a nice model for injective functions.

### Injective Covariant Reperesentable

The following property holds of covariant representables:

$$ℂ(S, h)$$ is injective iff $$h$$ is injective.

#### Proof

First we show that if $$h$$ is injective then $$ℂ(S, h)$$ is injective:

ℂ(S, h) f == ℂ(S, h) g
== {- definition ℂ(S, h) -}
h . f == h . g
=> {- injective h -}
f == g


Then we show that if $$ℂ(S, h)$$ is injective then $$h$$ is injective:

h . f == h . g
== {- definition ℂ(S, h) -}
ℂ(S, h) f == ℂ(S, h) g
=> {- injective ℂ(S, h) -}
f == g


### Injective Contravariant Reperesentable

Here’s a property of the contravariant representable functor:

$$ℂ(h, S)$$ is injective iff $$h$$ is surjective.

#### Proof

We work with the contrapositive to show that if $$h$$ is surjective, then $$ℂ(h, S)$$ is injective:

f ≠ g
== {- definition inequality -}
∃ y . f y ≠ g y
== {- surjective h -}
∃ x . f (h x) ≠ g (h x)
== {- definition inequality -}
f . h ≠ g . h
== {- definition ℂ(h, S) -}
ℂ(h, S) f ≠ ℂ(h, S) g


Finally we show that if $$h$$ is not surjective, then $$ℂ(h, S)$$ is not injective:

let h 0 = 0 {- h : {0} -> {0,1} -}
let f 0 = 0, f 1 = 0
let g 0 = 0, g 1 = 1


then

f . h = g . h
== {- definition ℂ(h, S) -}
ℂ(h, S) f = ℂ(h, S) g


but

 f ≠ g


Sadly, I don’t like this part of the proof, since it involves finding a counterexample to the injectivity of $$ℂ(h, S)$$ when $$h$$ is not surjective, and I prefer constructive proofs. Nevertheless, we’ve shown that $$ℂ(h, S)$$ is injective iff $$h$$ is surjective, which gives us the right cancellative definition of surjection.