category of paths on a graph

Let $G$ be an undirected graph. Denote the set of vertices of $G$ by “$V$” and denote the set of edges of $G$ by “$E$”.

A path of the graph $G$ is an ordered tuplet of vertices $(x_{1},x_{2},\ldots x_{n})$ such that, for all $i$ between $1$ and $n-1$, there exists an edge connecting $x_{i}$ and $x_{i+i}$. As a special case, we allow trivial paths which consist of a single vertex — soon we will see that these in fact play an important role as identity elements  in our category.

In our category, the vertices of the graph will be the objects and the morphisms  will be paths; given two of these objects $a$ and $b$, we set $\operatorname{Hom}(a,b)$ to be the set of all paths $(x_{1},x_{2},\ldots x_{n})$ such that $x_{1}=a$ and $x_{n}=b$. Given an object $a$, we set $1_{a}=(a)$, the trivial path mentioned above.

To finish specifying our category, we need to specify the composition operation. This operation  will be the concatenation of paths, which is defined as follows: Given a path $(x_{1},x_{2},\ldots,x_{n})\in\operatorname{Hom}(a,b)$ and a path $(y_{1},y_{2},\ldots,y_{m})\in\operatorname{Hom}(a,b)$, we set

 $a\circ b=(x_{1},x_{2},\ldots x_{n},y_{2},\ldots,y_{m}).$

(Remember that $x_{n}=y_{1}=b$.) To have a bona fide category, we need to check that this choice satisfies the defining properties (A1 - A3 in the entry http://planetmath.org/node/965category). This is rather easily verified.

A1: Given a morphism $(x_{1},x_{2},\ldots x_{n})$, it can only belong to $\operatorname{Hom}(a,b)$ if $x_{1}=a$ and $x_{n}=b$, hence $\operatorname{Hom}(a,b)\cup\operatorname{Hom}(c,d)=\emptyset$ unless $a=c$ and $b=d$.

A2: Suppose that we have four objects $a,b,c,d$ and three morphisms, $(x_{1},x_{2},\ldots x_{n})\in\operatorname{Hom}(a,b)$, $(y_{1},y_{2},\ldots y_{m})\in\operatorname{Hom}(b,c)$, and $(z_{1},z_{2},\ldots z_{k})\in\operatorname{Hom}(c,d)$. Then, by the definition of the operation $\circ$ given above,

 $\displaystyle((x_{1},x_{2},\ldots,x_{n})\circ($ $\displaystyle y_{1},y_{2},\ldots,y_{m}))\circ(z_{1},z_{2},\ldots,z_{k})$ $\displaystyle=(x_{1},x_{2},\ldots,x_{n},y_{2},\ldots,y_{m})\circ(z_{1},z_{2},% \ldots,z_{k})$ $\displaystyle=(x_{1},x_{2},\ldots,x_{n},y_{2},\ldots,y_{m},z_{2},\ldots,z_{k})$ $\displaystyle(x_{1},x_{2},\ldots,x_{n})\circ($ $\displaystyle(y_{1},y_{2},\ldots,y_{m})\circ(z_{1},z_{2},\ldots,z_{k}))$ $\displaystyle=(x_{1},x_{2},\ldots,x_{n})\circ(y_{1},y_{2},\ldots,y_{m},z_{2},% \ldots,z_{k})$ $\displaystyle=(x_{1},x_{2},\ldots,x_{n},y_{2},\ldots,y_{m},z_{2},\ldots,z_{k}).$

Since these two quantities are equal, the operation is associative.

A3: It is easy enough to check that paths with a single vertex act as identity elements:

 $\displaystyle(x_{1})\circ(x_{1},x_{2},\ldots,x_{n})$ $\displaystyle=(x_{1},x_{2},\ldots,x_{n})$ $\displaystyle(x_{1},x_{2},\ldots,x_{n})\circ(x_{n})$ $\displaystyle=(x_{1},x_{2},\ldots,x_{n})$

It is also possible to consider the equivalence class   of paths modulo retracing. To introduce this category, we first define a binary relation  $\approx$ on the class of paths as follows: Let $A$ and $B$ be any two paths such that the right endpoint  of $A$ is the same as the left endpoint of $B$, i.e. $A\in\operatorname{Hom}(a,b)$ and $B\in\operatorname{Hom}(b,c)$ for some vertices $a,b,c$ of our graph. Let $d$ be any vertex which shares an edge with $d$. Then we set $A\circ B\approx A\circ(c,d,c)\circ B$.

Let $\sim$ be the smallest equivalence relations which contains $\approx$. We will call this equivalence relation retracing.

As defined above, it may not intuitively obvious what this equivalence amounts to. To this end, we may consider a different description. Define the reversal of a path to be the path obtained by reversing the order of the vertices traversed:

 $(x_{1},x_{2},\ldots,x_{n-1},x_{n})^{-1}=(x_{n},x_{n-1},\ldots,x_{2},x_{1})$

Then we may show that two paths are equivalent    under retracing if they may both be obtained from a third path by inserting terms of the form $XX^{-1}$. In symbols, we claim that $A\sim B$ if there exists an integer $nï¼ž0$ and paths $X_{1},\ldots X_{n+1},Y_{1},\ldots Y_{n-1},Z_{1},\ldots Z_{n}$ such that

 $A=X_{1}\circ X_{1}^{-1}\circ Z_{1}\circ X_{2}\circ X_{2}^{-1}\circ\cdots\circ X% _{n-1}\circ X_{n-1}^{-1}\circ Z_{n}\circ X_{n}\circ X_{n}^{-1}\circ Z_{n}\circ X% _{n+1}\circ X_{n+1}^{-1}$

and

 $B=Y_{1}\circ Y_{1}^{-1}\circ Z_{1}\circ Y_{2}\circ Y_{2}^{-1}\circ\cdots\circ Y% _{n-1}\circ Y_{n-1}^{-1}\circ Z_{n}\circ Y_{n}\circ Y_{n}^{-1}\circ Z_{n}\circ Y% _{n+1}\circ Y_{n+1}^{-1}$

This characterization explains the choice of the term “retracing” — we do not change the equivalence class of the path if we happen to wander off somewhere in the course of following the path but then backtrack and pick the path up again where we left off on our digression.

Rather than presenting a detailed formal proof, we will sketch how the two definitions may be shown to be equivalent.

Title category of paths on a graph CategoryOfPathsOnAGraph 2013-03-22 16:45:54 2013-03-22 16:45:54 rspuzio (6075) rspuzio (6075) 18 rspuzio (6075) Example msc 20L05 msc 18B40 IndexOfCategories