With a clearer picture of determination and choice established in the last post, we now turn to refining the picture of the ‘division problem’ within the category of sets. The basic mechanic in both of these problems is trying to find the missing side of the triangle where the base is the direct mapping $h: A \rightarrow C$ and the two sides are the mappings $f:A \rightarrow B$ and $g:B \rightarrow C$ composed together, with the set $B$ acting as a waypoint. The only difference is between the two problems is which of the two sides is considered given and which is being solved for. In other words, the idea is to ‘divide’ both sides by $f$ to solve $f \circ g? = h$, in the determination case, or to ‘divide’ both sides by $g$ to solve $f? \circ g = h$ in the choice case, knowing full well that in certain circumstances an answer simply does not exist.

Lawvere and Schanuel, without explicitly saying clearly why in words, pay particular attention to a smaller subset of the determination and choice problems in which base of the triangle is an automorphism $I_A: A \rightarrow A$.  In this special case, the solutions to the determination and choice problems take the special name retractions and sections, respectively.

The two mathematical relationships that summarize these special cases are $ r \circ f = I$ and $f \circ s = I$ and where $f$ is the given mapping in whichever of the two problems is being considered (although, to be clear, $f$ will sometimes be called $g$ when it is the second leg of the composition and the context is clearer to do so). 

Of course, no matter which leg of the waypoint is specified as $f$, once a solution is found, both sides enter a dual relationship with the other and it then becomes a matter of taste as to which is the retraction and which is the section. Thus, $f$ can be regarded as the section to the retraction $r$ in the determination case and as the retraction to the section $s$ in the choice case. The following diagram illustrates this duality for a particular set of mappings between a two-element set (the label $A$ being suppressed for clarity) and the three-element set $B$.

The green lines correspond to the specific components of the mapping being solved for.  The dotted lines correspond to arrows that could be chosen differently (e.g. $b_2$ could point to either element in $A$). 

Once a section or a retraction have been found, other, more general ‘division’ problems can be solved.  In some sense, a retraction or a section are the primitives that unlock these problems.

The first of these general cases is built upon the retraction problem, where the mapping $y$, which goes from $A$ to a new set $T$, is given and the question is whether there exists a mapping $x$ from $B$ to $T$.  The ‘division problem’ we are trying to solve is defined by

\[ x \circ f = y \; . \]

The triangle portion of problem is summarized on the left where the ‘circular arrow’ connecting $A$ to $A$ is meant to remind us of the automorphism. 

Lawvere and Schanuel point out that this problem always has a solution when $x \equiv y \circ r$, since

\[ x \circ f = (y \circ r ) \circ f = y \circ ( r \circ f ) = y \circ I_A = y \; . \]

A specific instance of this, for the two- and three-element sets above, is given by the following diagram,

where the components of $y$ are the blue arrows, and the black and purple arrows are the original $f$ and $r$ dual pair of the retraction problem, respectively. The action of the retraction $r$ is to bring an element of $B$ back to $A$ where it can use the mapping $y$ to have access to the set $T$. The specific pieces of $x$ are:

\[ x(b_1) = (y \circ r) (b_1) = y (a_1) = t_1 \; , \]

\[ x(b_2) = (y \circ r) (b_2) = y (a_2) = t_4 \; , \]

and

\[ x(b_3) = (y \circ r) (b_3) = y (a_2) = t_4 \; . \]

In analogous way, the other, general cases is built upon the section problem, where the mapping $y$, which goes from a new set $T$ to $A$, is given and the question is whether there exists a mapping $x$ from $T$ to $B$. The ‘division problem’ we are now trying to solve is defined by

\[ f \circ x = y \; . \]

The triangle portion of problem is now summarized on the right where, again, the ‘circular arrow’ connecting $A$ to $A$ is meant to remind us of the automorphism. 

This problem always has a solution when $x \equiv s \circ y$, since

\[ f \circ x = f \circ (s \circ y) = (f \circ s) \circ y = I_A \circ y = y \; . \]

A specific instance of the section problem, again for the two- and three-element sets above, is given by the following diagram,

where the components of $y$ are still the blue arrows, and the black and green arrows are the original $f$ and $s$ dual pair of the section problem, respectively. The action of the section is to connect the action of the mapping $y: T \rightarrow $ to the set $B$ thereby putting each element of $T$ into the appropriate correspondence with elements in set $B$. The specific pieces of $x$ are:

\[ x(t_1) = (s \circ y) (t_1) = s (a_1) = b_1 \; , \]

\[ x(t_2) = (s \circ y) (t_2) = s (a_1) = b_1 \; , \]

\[ x(t_3) = (s \circ y) (t_3) = s (a_1) = b_1 \; , \]

and

\[ x(t_4) = (s \circ y) (t_4) = s (a_2) = b_3 \; . \]

One final note about the retraction and section problems. By making the base mapping of the triangle problem an automorphism, we are forces to supply additional requirements not imposed in the more general case. As was seen in the last post, an essential ingredient is that the left-side mapping from $A$ to $B$ be injective and the right-side mapping from $B$ to $A$ be surjective. It is instructive to look at how the number of retractions vary with size of the waypoint set $B$. The following diagram illustrates the retraction problem for the three cases with either one fewer, the same number of, or one more elements in $B$ than in $A$ (where $f$ and $g$ are used to specify the left- and right-side of the triangle).

Likewise, the following diagram illustrates how the number of sections vary through the same progression in the set $B$.

In both cases, there is a sense that the size of $B$ (i.e.  the number of the elements) is central to whether or not the problem has a solution and, if so, how many solutions exist.  This observation leads to two conclusions.

First, the definition of an inverse mapping, which was originally specified by the two conditions $g \circ f = I_A$ and $f \circ g = I_B$ (see this post for details), can now be recast as:

A map $f:A \rightarrow B$ is an isomorphism if there exists a unique inverse map $f^{-1}$ which is both a retraction and a section for $f$ such that $f \circ f^{-1} = I_B$ and $f^{-1} \circ f = I_A$.

Since the inverse must be both injective and surjective (since it is both a retraction and a section), we arrive at the familiar result that an inverse must be bijective. 

Second, one may wonder why all this machinery is needed to arrive at a result already familiar to mathematics centuries earlier.  According to Lawver and Schanuel, all this hard work will pay off when tackling categories where the objects are promoted from simple finite sets to richer sorts.  In particular, infinite sets and dynamical systems are cited but the jury is still out in this exploration as to whether their argument will ultimately be convincing.