Over a cocomplete base, a monadic category is cocomplete exactly when it has coequalizers

For the forward direction, if $\mathcal{A}$ is cocomplete then it has the colimit of every parallel pair, hence every coequalizer.

For the reverse direction, replace $\mathcal{A}$ across its monadic comparison equivalence by ${\mathcal{C}}^T$. Given a small family $(A_i,a_i)$, [L5] and [L6] identify the coproducts of free algebras $P:={\coprod }_i{F}^T(T{A}_i),Q:={\coprod }_i{F}^T(A_i)$ with free algebras on the corresponding base coproducts. If $j_i$ and $k_i$ are their coproduct injections, define $\alpha ,\beta :P\rightrightarrows Q$ by $\alpha j_i=k_{i}T(a_i),\beta j_i=k_i{\mu }_{A_i}\mathrm{.}$ By hypothesis this pair has a coequalizer $q:Q\to R$ in ${\mathcal{C}}^T$. The construction also applies to the empty family.

For every $i$, the map $q{k}_i$ coequalizes the canonical pair for $(A_i,a_i)$. Its universal property [L4] gives a unique algebra map ${\iota }_i:(A_i,a_i)\to R$ satisfying ${\iota }_i{a}_i=q{k}_i$.

Given algebra maps $r_i:(A_i,a_i)\to (D,d)$, the maps $r_i{a}_i:F^T(A_i)\to (D,d)$ assemble to a map $r:Q\to (D,d)$. Each coequalizes its canonical pair by [L4], so $r\alpha =r\beta$ and there is a unique $\bar{r}:R\to (D,d)$ with $\bar{r}q=r$. Then $\bar{r}{\iota }_i=r_i$ because the two maps agree after the epimorphism $a_i$, and uniqueness follows because the $q{k}_i={\iota }_i{a}_i$ are jointly epimorphic. Thus $(R,({\iota }_i))$ is the coproduct of the family.

For the reverse direction, ${\mathcal{C}}^T$ now has all small coproducts by step 3.1 and all coequalizers by hypothesis, so [L2] constructs every small colimit.

For the reverse direction, transport these colimits back across the comparison equivalence by [L3]. Thus $\mathcal{A}$ is cocomplete.

Step 1.1 proves the forward implication, while steps 1.2 to 5.1 prove the reverse implication, establishing the biconditional.