Vtyjkyuk

Please tell me how to solve the following exercise which is in the book "Conceptual Mathematics A first introduction to categories Second Edition".

How can I show that for any object X of category of graphs the diagram is commutative from the fact that following two diagrams are commutative?

The first observation to make is that given $B_1overseta_1leftarrow Coverseta_2rightarrow B_2$, a morphism $Ato C$ (resp. $Dto C$) gives rise to a pair of morphisms $B_1leftarrow Arightarrow B_2$ (resp. $B_1leftarrow Drightarrow B_2$) by composition with the $a_i$. And by hypothesis, such pairs correspond to morphisms $Ato P$ (resp. $Dto P$). If there are morphisms $f,g:Cto P$ with $$pi_1circ f=pi_1circ g=a_1$$ and $$pi_2circ f=pi_2circ g=a_2,$$ then for any $x:Ato C$ it's easy to check that $fcirc x=gcirc x$ by $P$ being a "product with respect to $A$ and $D$", and similarly for any morphism $Dto C$; but then $f=g$ by the (italicized in the text) property of morphisms from $A$ or $D$. So if there is a morphism $f:Cto P$ making $pi_icirc f=a_i$, it is unique.

How do we prove the existence of such a unique $f$? I don't believe the authors have mentioned the really nice thing yet about this category of graphs, where every object is actually a colimit of a diagram consisting of $A$s and $D$s; from this, existence is almost immediate. But you can figure it out element-wise by taking an edge $e$ of $C$ to $langle a_1circ tildee,a_2circ tildeerangle(e_0)$, where $tildee$ is the morphism $Ato C$ that picks out the edge $e$, $e_0$ is the unique edge of $A$, and $langle a_1circ tildee,a_2circ tildeerangle$ is the unique morphism $Ato P$ corresponding to the pair of morphisms $a_icirctildee:Ato B_i$ (and you do the essentially thing for the graph's vertices and morphisms $Dto C$).

All that's needed is to confirm that this assignment really is a morphism of graphs, which is boring but not too hard. It should be clear that the process I've just described gives us a function from the edges of $C$ to the edges of $P$, and likewise for the vertices. All that's needed to check that it's a morphism of graphs is to check that $$s_P(langle a_1circtildee,a_2circtildeerangle(e_0))=langle a_1circwidetildes_C(e),a_2circwidetildes_C(e)rangle(d_0)$$ and likewise $$t_P(langle a_1circtildee,a_2circtildeerangle(e_0))=langle a_1circwidetildet_C(e),a_2circwidetildet_C(e)rangle(d_0),$$ where $d_0$ and $s_ldots$ and $t_ldots$ are the source and target operations on the subscripted graphs. What will make this easier to check is that if $d=s_C(e)$, then, where $x_1,x_2:Dto A$ are the morphisms picking out the source and target vertices respectively in $A$, then $$widetildes_C(e)=tilded=tildeecirc x_1$$ and $$widetildet_C(e)=tildeecirc x_2.$$ Moreover $$s_P(langle a_1circtildee,a_2circtildeerangle(e_0))=langle a_1circtildee,a_2circtildeeranglecirc x_1(d_0),$$ and similarly with $t$ and $x_2$. With these equalities it should be easy to check.