In the following link <http://www.math.jhu.edu/~eriehl/ssets.pdf> on page 8 there is the following diagram
$$\begin{array}
XX\_m .\mathcal{F}[n] & \stackrel{f\_{\*}} {\longrightarrow} & X\_m.\mathcal{F}[m] \\
\downarrow{f^{\*}} & & \downarrow{\gamma\_m} \\
X\_n. \mathcal{F}[n] & \stackrel{\gamma\_n}{\longrightarrow} & e \thinspace , 
\end{array}
$$ 


And defines the object $e$ along with the morphisms $\gamma\_n$ which make the diagram commutative for each choice of $f$ to be called wedge. Now calls a universal such an edge (which seems to be *like* the pushout of the diagram, but with the additional assumption of being universal with respect any choice of $f's$) coend and denotes it by this *"fancy"* (strange) integral notation $\int^{n} X\_n . \mathcal{F}[n]$ (for which I think another MOF question exists, with a very good answer, however whoever wants to mention something about it is more than welcome). After that though, there is a tricky equivalence which I don't really get and your help is probably necessary. Says, that equivalently the coend can be defined as the coequalizer of a certain diagram explicitly written below that sentence. So my question is why is it equivalent, or in other words, how do we prove that? 


Should be kind of obvious probably, but I'm not very comfortable at the moment with those things hence isn't for me.


If instead of an answer/comment someone has a reference, then they are really welcome!



Coend as a universal edge
Try to have a look to section 1.2 of [this](https://arxiv.org/pdf/1501.02503.pdf), and feel free to comment with help requests!


