Let $A$, $B$, and $C$ be sets.

  1. Associativity. We have isomorphisms of sets[1]
    \[ \underbrace{\text{CoEq}\webleft (\text{coeq}\webleft (f,g\webright )\circ f,\text{coeq}\webleft (f,g\webright )\circ h\webright )}_{{}=\text{CoEq}\webleft (\text{coeq}\webleft (f,g\webright )\circ g,\text{coeq}\webleft (f,g\webright )\circ h\webright )}\cong \text{CoEq}\webleft (f,g,h\webright ) \cong \underbrace{\text{CoEq}\webleft (\text{coeq}\webleft (g,h\webright )\circ f,\text{coeq}\webleft (g,h\webright )\circ g\webright )}_{{}=\text{CoEq}\webleft (\text{coeq}\webleft (g,h\webright )\circ f,\text{coeq}\webleft (g,h\webright )\circ h\webright )}, \]

    where $\text{CoEq}\webleft (f,g,h\webright )$ is the colimit of the diagram

    in $\mathsf{Sets}$.

  2. Unitality. We have an isomorphism of sets
    \[ \text{CoEq}\webleft (f,f\webright )\cong B. \]
  3. Commutativity. We have an isomorphism of sets
    \[ \text{CoEq}\webleft (f,g\webright ) \cong \text{CoEq}\webleft (g,f\webright ). \]
  4. Interaction With Composition. Let
    \[ A \underset {g}{\overset {f}{\rightrightarrows }} B \underset {k}{\overset {h}{\rightrightarrows }} C \]

    be functions. We have a surjection

    \[ \text{CoEq}\webleft (h\circ f,k\circ g\webright )\twoheadrightarrow \text{CoEq}\webleft (\text{coeq}\webleft (h,k\webright )\circ h\circ f,\text{coeq}\webleft (h,k\webright )\circ k\circ g\webright ) \]

    exhibiting $\text{CoEq}\webleft (\text{coeq}\webleft (h,k\webright )\circ h\circ f,\text{coeq}\webleft (h,k\webright )\circ k\circ g\webright )$ as a quotient of $\text{CoEq}\webleft (h\circ f,k\circ g\webright )$ by the relation generated by declaring $h\webleft (y\webright )\sim k\webleft (y\webright )$ for each $y\in B$.

Item 1: Associativity
Omitted.
Item 2: Unitality
Clear.
Item 3: Commutativity
Clear.
Item 4: Interaction With Composition
Omitted.


Footnotes

[1] That is, the following three ways of forming “the” coequaliser of $\webleft (f,g,h\webright )$ agree:
  1. Take the coequaliser of $\webleft (f,g,h\webright )$, i.e. the colimit of the diagram
    in $\mathsf{Sets}$.
  2. First take the coequaliser of $f$ and $g$, forming a diagram
    \[ A\underset {g}{\overset {f}{\rightrightarrows }}B\overset {\text{coeq}\webleft (f,g\webright )}{\twoheadrightarrow }\text{CoEq}\webleft (f,g\webright ) \]
    and then take the coequaliser of the composition
    \[ A\underset {h}{\overset {f}{\rightrightarrows }}B\overset {\text{coeq}\webleft (f,g\webright )}{\twoheadrightarrow }\text{CoEq}\webleft (f,g\webright ), \]
    obtaining a quotient
    \[ \text{CoEq}\webleft (\text{coeq}\webleft (f,g\webright )\circ f,\text{coeq}\webleft (f,g\webright )\circ h\webright )=\text{CoEq}\webleft (\text{coeq}\webleft (f,g\webright )\circ g,\text{coeq}\webleft (f,g\webright )\circ h\webright ) \]
    of $\text{CoEq}\webleft (f,g\webright )$
  3. First take the coequaliser of $g$ and $h$, forming a diagram
    \[ A\underset {h}{\overset {g}{\rightrightarrows }}B\overset {\text{coeq}\webleft (g,h\webright )}{\twoheadrightarrow }\text{CoEq}\webleft (g,h\webright ) \]
    and then take the coequaliser of the composition
    \[ A\underset {g}{\overset {f}{\rightrightarrows }}B\overset {\text{coeq}\webleft (g,h\webright )}{\twoheadrightarrow }\text{CoEq}\webleft (g,h\webright ), \]
    obtaining a quotient
    \[ \text{CoEq}\webleft (\text{coeq}\webleft (g,h\webright )\circ f,\text{coeq}\webleft (g,h\webright )\circ g\webright )=\text{CoEq}\webleft (\text{coeq}\webleft (g,h\webright )\circ f,\text{coeq}\webleft (g,h\webright )\circ h\webright ) \]
    of $\text{CoEq}\webleft (g,h\webright )$.

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