Contents

# Contents

## Definition

A function $f$ (of sets) from $A$ to $B$ is surjective if, given any element $y$ of $B$, $y = f(x)$ for some $x$. A surjective function is also called onto or a surjection.

Surjections are the same as epimorphisms in the category of sets and effective epimorphisms in the (infinity,1)-category of sets.

A bijection is a function that is both surjective and injective.

## The surjection preorder

In classical set theory, one writes $|B| \leq^* |A|$ to mean that either there is a surjection $A \to B$ or $B=\empty$. The relation $\leq^*$ is a preorder on the class of all sets, and its restriction to inhabited sets is the preorder reflection of the category $Surj_{inh}$ of inhabited sets and surjections.

To make this definition less piecemeal and more constructive, one can define $|B| \leq^* |A|$ to mean $B$ is a subquotient of $A$, in other words one has a surjection $B' \to B$ and an injection $B' \to A$. For $B$ inhabited and a subquotient of $A$, excluded middle implies there is a surjection from $A$ to $B$, so in classical mathematics this coincides with the piecemeal definition.

Contrast with the notation $|B| \leq |A|$ if there is an injection $B\to A$. Since subobjects are subquotients (e.g. we can take $B'=B$ above), $|B| \leq |A|$ implies $|B| \leq^* |A|$. (If we wanted to prove this using the piecemeal definition, we would require excluded middle.)

## Axioms of choice

The axiom of choice states precisely that every surjection in the category of sets has a section. Thus in this setting one has: $|B| \leq^* |A|$ implies $|B| \leq |A|$, and so $|B| \leq^* |A|$ iff $|B| \leq |A|$ assuming AC. Some authors who doubt the axiom of choice use the term ‘onto’ for a surjection as defined above and reserve ‘surjective’ for the stronger notion of a function with a section (a split epimorphism).

The axiom WISC has an equivalent statement (that works in any Boolean topos) due to François Dorais phrased almost entirely in terms of surjections (or epimorphisms):

For every set $X$ there is a set $Y$ such that for every surjection $q\colon Z \to X$ there is a function $s\colon Y \to Z$ such that $q\circ s\colon Y\to X$ is a surjection.

One can view this as really a statement about the Grothendieck fibration over Set with fibre over $X$ the full subcategory of $Set/X$ on the surjections: every fibre has a weakly initial object.

## In other categories

Since an element $a$ in a set $A$ in the category of sets is just a global element $a:1\rightarrow A$, one could define surjections in any category $\mathcal{C}$ with a terminal object $1$:

###### Definition

A morphism $f:A\rightarrow B$ in a category $\mathcal{C}$ with a terminal object $1$ is called a surjection. a surjective morphism, or an onto morphism if, given any global element $y:1\rightarrow B$, there exists a global element $x:1\rightarrow A$ such that $y = f \circ x$.

###### Remark

Some authors regard surjection as a synonym of split epimorphism, and only use ‘onto’ for the definition above.

###### Proposition

In a category $\mathcal{C}$ with a terminal object $1$, the unique morphism $!:A\rightarrow 1$ is a surjection for every object $A$.

###### Proof

By definition of a terminal object, for every object $A$ there exists a unique morphism $!:A\rightarrow 1$, and the identity morphism is the unique global element $1_{1}:1\rightarrow 1$. The composite of a global element $x:1\rightarrow A$ with the function $!:A\rightarrow 1$ results in a function $! \circ x:1\rightarrow 1$, which by definition of a terminal object is the same as $1_{1}:1\rightarrow 1$. Since $! \circ x = 1_{1}$, for every object $A$, $!:A\rightarrow 1$ is a surjection.

###### Proposition

In a category of pointed objects $\mathcal{C}$, every morphism $f:A\rightarrow B$ is a surjection for every object $A$.

###### Proof

A pointed object in a category with terminal objects is a object $A$ with a global element $a:1\rightarrow A$ to the object, and morphisms in the category preserve the global element, which means there is only a unique global element for each object $A$. Therefore, for every morphism $f:A\rightarrow B$ and global element $y:1\rightarrow B$, there exists a global element $x:1\rightarrow A$ such that $y = f \circ x$.

Just because a morphism is a surjection in a concrete category does not mean that the morphism in the underlying category of sets is a surjection; equivalently, the forgetful functor from any category $\mathcal{C}$ to Set does not preserve surjections.

### Well-pointedness

The fact that surjections are epimorphisms in Set is a result of the fact that Set is well-pointed. This could be generalised to any category with a terminal separator $1$.

###### Proposition

In a category $\mathcal{C}$ with a terminal object $1$ such that $1$ is a separator, every surjection is an epimorphism.

###### Proof

For any surjection $f:A\rightarrow B$, suppose there are parallel morphisms $g, h:B\rightarrow C$ such that there $g \circ f = h \circ f$ (a fork). Then for every global element $y:1\rightarrow B$ there exists a global element $x:1\rightarrow A$ such that $y = f \circ x$, and thus $g \circ y = g \circ f \circ x$ and $h \circ y = h \circ f \circ x$. But since $g \circ f = h \circ f$, $g \circ f \circ x = h \circ f \circ x$, which implies that $g \circ y = h \circ y$. Since $1$ is a separator, then for every global element $y:1\rightarrow B$, if $g \circ y = h \circ y$, then $g = h$. Therefore, every surjection is an epimorphism.

### Axiom of choice

The axiom of choice for surjections in Set is the following statement:

This axiom could be defined in every category with a terminal object, and could be contrasted with the axiom of choice for epimorphisms, as not every epimorphism is an surjection, nor is every surjection an epimorphism.

###### Proposition

In a category $\mathcal{C}$ with a terminal object $1$ and binary equalisers such that every surjection is a split epimorphism, the terminal object $1$ is a separator.

###### Proof

Suppose there are parallel morphisms $g, h:B\rightarrow C$ such that for every global element $y:1\rightarrow B$, $g \circ y = h \circ y$. One could construct an equaliser $f:eq(g,h)\rightarrow B$, which implies that $g \circ f = h \circ f$ and that for any global element $x:1\rightarrow eq(g,h)$, $g \circ f \circ x = h \circ f \circ x$. This implies that for every global element $y:1\rightarrow B$, there exists a global element $x:1\rightarrow eq(g,h)$ where $y = f \circ x$, and the equaliser $f:eq(g,h)\rightarrow B$ is a surjection. Since every surjection is a split epimorphism, $f$ has a section $i:B\rightarrow eq(g,h)$ such that $f \circ i = 1_{B}$, the identity morphism on $B$, $g \circ f \circ i = h \circ f \circ i$, $g \circ 1_{B} = h \circ 1_{B}$, and $g = f$. Because for every global element $y:1\rightarrow B$, $g \circ y = h \circ y$ implies $g = h$, the terminal object $1$ is a separator.

Last revised on January 7, 2023 at 05:31:52. See the history of this page for a list of all contributions to it.