Contents

# Contents

## Definition

A binary relation from a set $X$ to a set $Y$ is called functional if every element of $X$ is related to at most one element of $Y$. Notice that this is the same thing as a partial function, seen from a different point of view. A (total) function is precisely a relation that is both functional and entire.

## Properties

Like any relation, a functional relation $r$ can be viewed as a span

$\array { & & \Delta_r \\ & \swarrow_{\iota_r} & & \searrow^{\phi_r} \\ X & & & & Y }$

Such a span is a relation iff the pairing map from the domain $\Delta_r$ to $X \times Y$ is an injection, and such a relation is functional iff the inclusion map $\iota_r$ is also an injection. Such a relation is entire iff the inclusion map $\iota_r$ is a surjection.

(Total) functions are precisely left adjoints in the bicategory of relations, in other words relations $r: X \to Y$ in $Rel$ for which there exists $s: Y \to X$ satisfying

$r \circ s \leq 1_Y; \qquad 1_X \leq s \circ r.$

For if a relation $r: X \to Y$ is a function, then

1. $r r^{op} \leq 1_Y$: this just says that if $r^{op}(y, x)$ and $r(x, y')$, then $y = y'$. Equivalently, if $r(x, y)$ and $r(x, y')$ then $y = y'$: this holds precisely because $r$ is functional or well-defined.

2. $1_X \leq r^{op} r$: this says that if $x = x'$ in $X$, then there exists $y$ in $Y$ such that $r(x, y)$ and $r^{op}(y, x')$, i.e., there exists $y$ such that $r(x, y)$ and $r(x', y)$. This holds precisely because a function is an entire relation.

On the other hand, suppose $r: X \to Y$ is a left adjoint.

1. From $1_X \leq s r$ we deduce that if $x = x'$ in $X$, then there exists $y$ in $Y$ such that $r(x, y)$ and $s(y, x')$; in particular $r$ is entire.

2. Suppose $r(x, y)$. From $x = x$ there exists $y'$ such that $r(x, y')$ and $s(y', x)$. Thus $s(y', x)$ and $r(x, y)$; from $s r \leq 1_Y$ we infer $y' = y$. We conclude at most one $y$ satisfies $r(x, y)$, making $r$ functional.

Since $r$ is a function, it has a right adjoint $r$, and by uniqueness of right adjoints, we may conclude $s = r^{op}$. The monad $t = r^{op} r$ has a unit $1_X \leq t$ ($t$ is reflexive) and a multiplication $t t \leq t$ ($t$ is transitive), and also $t^{op} = (r^{op} r)^{op} = r^{op} r^{op\; op} = r^{op} r = t$ ($t$ is symmetric). So the monad $t$ is an equivalence relation. The above reasoning may be internalized to apply to the bicategory of relations internal to a regular category, in which case this equivalence relation is exactly the kernel pair of the map $r$. The comonad $c = r r^{op}$ has a counit $c \leq 1_Y$ ($c$ is coreflexive), and such coreflective relations in $Set$ correspond to subsets of $Y$. More generally, in a regular category, the subobject of $Y$ named by $c$ is the image of $r$ (the coequalizer of the kernel pair).

Further remarks: surjectivity of a function $r: X \to Y$ can be expressed as the condition $1_Y \leq r \circ r^{op}$, and injectivity as $r^{op} \circ r \leq 1_X$.