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Revision difference : commutative diagram
Version 4 Version 3
\begin{defn} \begin{defn}
Let $\mathcal{C}$ be a category. A \emph{diagram} in $\CC$ is a Let $\mathcal{C}$ be a category. A \emph{diagram} in $\CC$ is a
directed graph $\GG$ with vertex set $V$ and edge set $E$, (``loops'' directed graph $\GG$ with vertex set $V$ and edge set $E$, (``loops''
and ``parallel edges'' are allowed) together with two maps and ``parallel edges'' are allowed) together with two maps
$o\co V\to\mathrm{Obj}(\CC)$, $m\co E\to \mathrm{Morph}(\CC)$ such that $o\co V\to\mathrm{Obj}(\CC)$, $m\co E\to \mathrm{Morph}(\CC)$ such that
if $e\in E$ has source $s(e)\in V$ and target $t(e)\in V$ then if $e\in E$ has source $s(e)\in V$ and target $t(e)\in V$ then
$m(e) \in \text{Hom}_{\CC}\left(s(e),t(e)\right)$. $m(e) \in \text{Hom}_{\CC}\left(s(e),t(e)\right)$.
\end{defn} \end{defn}
Usually diagrams are denoted by drawing the corresponding graph Usually diagrams are denoted by drawing the corresponding graph
and labeling its vertices (respectively edges) with their images under $o$ and labeling its vertices (respectively edges) with their images under $o$
(respectively $m$), for example if $f\co A\to B$ is a morphism (respectively $m$), for example if $f\co A\to B$ is a morphism
$$\xymatrix@1{ {A}\ar[r]^f&{B} }$$ $$\xymatrix@1{ {A}\ar[r]^f&{B} }$$
is a diagram. Often (as in the previous example) the vertices themselves are is a diagram. Often (as in the previous example) the vertices themselves are
not drawn since their position can b deduced by the position of their not drawn since their position can b deduced by the position of their
labels. labels.
\begin{defn} \begin{defn}
Let $D=(\GG,o,m)$ be a diagram in the category $\CC$ and $\Gg=(e_1,\ldots,e_n)$ Let $D=(\GG,o,m)$ be a diagram in the category $\CC$ and $\Gg=(e_1,\ldots,e_n)$
be a path in $\GG$. Then the \emph{composition along} $\Gg$ is the following be a path in $\GG$. Then the \emph{composition along} $\Gg$ is the following
morphism of $\CC$ morphism of $\CC$
$$\circ(\Gg):=m(e_n)\circ\cdots\circ m(e_1)\,.$$ $$\circ(\Gg):=m(e_n)\circ\cdots\circ m(e_1)\,.$$
We say that $D$ is We say that $D$ is
\emph{commutative} or that it \emph{commutes} if for any two objects in \emph{commutative} or that it \emph{commutes} if for any two objects in
the image of $o$, say $A=o(v_1)$ and $B=o(v_2)$, and any two paths $\Gg_1$ the image of $o$, say $A=o(v_1)$ and $B=o(v_2)$, and any two paths $\Gg_1$
and $\Gg_2$ that connect $v_1$ to $v_2$ we have and $\Gg_2$ that connect $v_1$ to $v_2$ we have
$$\circ(\Gg_1)=\circ(\Gg_2)\,.$$ $$\circ(\Gg_1)=\circ(\Gg_2)\,.$$
\end{defn} \end{defn}
For example the commutativity of the triangle For example the commutativity of the triangle
$$\xymatrix{ $$\xymatrix{
{A}\ar[rr]^{f}\ar[dr]_{h}&&{B}\ar[dl]^{g}\\ {A}\ar[rr]^{f}\ar[dr]_{h}&&{B}\ar[dl]^{g}\\
&{C}& &{C}&
translates to $h=g\circ f$, while the commutativity of the square translates to $h=g\circ f$, while the commutativity of the square
$$\xymatrix{ $$\xymatrix{
{A}\ar[r]^{f}\ar[d]_{k}&{B}\ar[d]^{g}\\ {A}\ar[r]^{f}\ar[d]_{k}&{B}\ar[d]^{g}\\
{C}\ar[r]_{h}&{D} {C}\ar[r]_{h}&{D}
translates to $g\circ f=h\circ k$. translates to $g\circ f=h\circ k$.