What are the inclusion arrows in the coproducts of the category of algebras for a monad? $\newcommand{\A}{\mathscr{A}}\newcommand{\C}{\mathsf{C}}\newcommand{\T}{\mathcal{T}}\newcommand{\id}{\operatorname{id}}$Riehl, proposition $5.6.11$, from Category Theory in Context:

Suppose $\C$ is a cocomplete category and $\A$ is a category with all coequalisers. If $\A$ is monadic over $\C$, then $\A$ is cocomplete.

The proof is mostly left as an exercise. If you are unfamiliar with the notation used, I explain it in the second section. There is a very similar nLab article that is equally brief. I'll paraphrase her sketch:

By cocontinuity of equivalences, we can replace $\A$ by the category $\C^\T$ where $\T$ is the relevant monad, and $\C^\T$ must have all coequalisers.
Let $(A_i,\alpha_i)_{i\in I}\in\C^\T$ be any given family of algebras. The coproduct $(A,\alpha)$ of this family is defined to be the coequaliser at the right of the following diagram:



All objects in the fork are free $\T$-algebras. Using the adjunction $U^\T\vdash F^\T$ it is straightforward to show $(A,\alpha)$ is a coproduct. Then $\C^\T$ has all coproducts and coequalisers, thus all colimits.

Here, I assume $\kappa:\bigsqcup_i\T(A_i)\to \T(\bigsqcup_i A_i)$ is the unique arrow induced by the arrows $\T(j):\T(A_j)\to\T(\bigsqcup_i A_i)$, $j:A_j\hookrightarrow\bigsqcup_iA_i$ being the coproduct inclusion.
The trouble is, I can't make out what the inclusions arrows $(A_i,\alpha_i)\hookrightarrow (A,\alpha)$ are supposed to be! The nLab article says that it is not difficult to show $A=\bigsqcup_i A_i$, but what should $\alpha$ be? If we can construct this coequaliser by hand anyway, what's the point of assuming coequalisers exist? It just doesn't make sense to me.
My thoughts:
$(A,\alpha)$ is created as a coequaliser. We know nothing about it other than this, so to get an arrow $(A_j,\alpha_j)\to(A,\alpha)$ I have to route it through any of the objects in the coequaliser fork, almost certainly the free algebra $F^\T(\bigsqcup_i A_i)$. However, finding arrows into $F^\T$ is hard, since it is the left adjoint.
I need to find some arrow $\iota_j:A_j\to\T(\bigsqcup_i A_i)$ for every $j$ and: (i) hope $\iota_j$ lifts to an algebra homomorphism, (ii) hope that it is the right choice of inclusion arrow for the coproduct universal property. There is only one 'natural' choice of $\iota_j$ that I can see: $$\large\left(A_j\overset{j}{\hookrightarrow}\bigsqcup_i A_i\overset{\eta_{\sqcup_i A_i}}{\longrightarrow}\T\left(\bigsqcup_i A_i\right)\right)=\left( A_j\overset{\eta_{A_j}}{\longrightarrow}T(A_j)\overset{T(j)}{\longrightarrow}T\left(\bigsqcup_i A_i\right)\right)$$
But unfortunately this isn’t necessarily a homomorphism. If this $\iota_j$ satisfies $\iota_j\alpha_j=\mu_{\sqcup_i A_i}\T(\iota_j)$, then that is iff: $$\begin{align}\T(j)\eta_{A_j}\alpha_j&=\mu_{\sqcup_i A_i}\T^2(j)\T(\eta_{A_j})\\&=\T(j)\mu_{A_j}\T(\eta_{A_j})\\&=\T(j)\end{align}$$Though $\alpha_j\eta_{A_j}=\id_j$, $\eta_{A_j}\alpha_j\neq\id_{\T(A_j)}$ in general. So this map isn't, as far as I can tell, guaranteed to be a homomorphism. There are also no other 'natural' choices, I think, for these inclusions - so I am stuck. By ‘natural’ I really mean to say: there just aren’t any other maps we can guarantee to exist, that I can think of.
I would really appreciate a concrete description of what the inclusion homomorphisms $(A_j,\alpha_j)\to(A,\alpha)$ are, and any hints / answers for showing $(A,\alpha)$ is indeed a coproduct would be most welcome.
Another point: as a left adjoint, $F^\T$ preserves all colimits, and $\C$ is cocomplete. Perhaps that is necessary here... but so far, Riehl has only suggested that we require $\C$ to have all coproducts - I'm not sure why $\C$ needs to have all colimits.

In case Riehl's notation isn't well-known, here is the definition of $\C^\T$:

For a monad $(\T,\eta,\mu)$ on a category $\C$, we define the Eilenberg-Moore category, a.k.a. category of $\T$-algebras, $\C^\T$ to have objects pairs $(c,\varsigma)$ where $c\in\C$ and $\varsigma:\T\C\to\C$ is some arrow in $\C$, the 'algebra structure map', which must satisfy the two conditions: $$\varsigma\eta_c=\id_c,\quad\varsigma\mu_c=\varsigma\T(\varsigma)$$Arrows in $\C^\T$ shall be "$\T$-algebra homomorphisms" $f:(c,\varsigma)\to(c',\varsigma')$ corresponding to arrows $f:c\to c'$ in $\C$ that satisfy the condition: $$f\varsigma=\varsigma'\T(f)$$The identity arrows and arrow composition are inherited from $\C$.
The adjunction she refers to is this: $F^\T:\C\to\C^\T$ is the functor assigning $c\mapsto(Tc;\mu_c)$ and $f:c\to c'$ is mapped to $T(f)$ ($F^\T$ creates 'free algebras' and 'free maps'). $U^\T:\C^\T\to\C$ is the forgetful functor.

 A: $\newcommand{\A}{\mathscr{A}}\newcommand{\C}{\mathsf{C}}\newcommand{\T}{\mathcal{T}}\newcommand{\id}{\operatorname{id}}$A less opaque exposition is given here but it still omits some details that caused me a headache or two... A key point is my last observation that $F^\T$ preserves colimits. I will explain their exposition here for posterity and also for my own note-taking.
In their notation, we consider a family $(X_i,\xi_i)_{i\in I}$ of algebras. We know that $\sum_i X_i$ exists in $\C$, and that: $$\left(\T\left(\sum_i X_i\right),\mu_{\sum_i X_i}\right)=F^\T\left(\sum_i X_i\right)\cong\sum_i(\T X_i,\mu_{X_i})$$Define the inclusions $j:X_j\to\sum_i X_j$, $j':(\T X_j,\mu_{X_j})\to\sum_i(\T X_i,\mu_{X_i})$, $j'':(\T^2 X_j,\mu_{\T X_j})\to\sum_i(\T^2 X_i,\mu_{\T X_i})$. We know that $T(j)$ corresponds to $j'$ under the isomorphism. This is because $F^\T$ is cocontinuous!
Now, we define an algebra $(X,\xi)$ by the coequaliser: $$\sum_i(\T^2 X_i,\mu_{\T X_i})\overset{\sum_i\mu_{X_i}}{\underset{\sum_i\T\xi_i}{\large\rightrightarrows}}\sum_i(\T X_i,\mu_{X_i})\cong\left(\T\left(\sum_i X_i\right),\mu_{\sum_i X_i}\right)\overset{p}\longrightarrow(X,\xi)$$
We define the inclusions as follows:

Set $h_j=p\circ\eta_{\sum_i X_i}\circ j:X_j\to X$. This is an algebra homomorphism because $p$ is (my oversight before was to not try to show $h_j$ a homomorphism directly, but to attempt to show each component is a homomorphism - which is false): $$\begin{align}\xi\circ\T(h_j)&=\xi\circ\T(p)\circ\T(\eta_{\sum_i X_i})\circ\T(j)\\&=p\circ\mu_{\sum_i X_i}\circ\T(\eta_{\sum_i X_i})\circ\T(j)\\&=p\circ\T(j)\\h_j\circ\xi_j&=p\circ\T(j)\circ\T(\xi_j)\circ\eta_{\T X_j}\\&=p\circ\T(j)\circ\mu_{X_i}\circ\eta_{\T X_j}\\&=p\circ\T(j)\end{align}$$Where the fact that $p$ is a homomorphism is used, but also that $p$ commutes with the parallel coproduct arrows $\sum_i\mu_{X_i},\,\sum_i\T(\xi_i)$.

So that settles my original question.
Now suppose $(\sigma,\varsigma)$ is some other $\T$-algebra with given homomorphisms $c_i:(X_i,\xi_i)\to(\sigma,\varsigma)$. Then $c_i\xi_i:(\T X_i,\mu_{X_i})\to(\sigma,\varsigma)$ are homomorphisms, inducing a unique homomorphism: $$g:\left(\T\left(\sum_i X_i\right),\mu_{\sum_i X_i}\right)\to(\sigma,\varsigma)$$For which $g\circ\T(j)=c_j\circ\xi_j$ for all $j\in I$. Then: $$\sum_i(\T^2 X_i,\mu_{\T X_i})\overset{\sum_i\mu_{X_i}}{\underset{\sum_i\T\xi_i}{\large\rightrightarrows}}\sum_i(\T X_i,\mu_{X_i})\cong\left(\T\left(\sum_i X_i\right),\mu_{\sum_i X_i}\right)\overset{g}\longrightarrow(\sigma,\varsigma)$$Commutes; it is equivalent to check commutativity on every coordinate. Precomposing with an inclusion $j''$ gives $g\T(j)\mu_{X_j}$ versus $g\T(j)\T(\xi_j)$, that is, $c_j\circ\xi_j\circ\mu_{X_j}$ versus $c_j\circ\xi_j\circ\T(\xi_j)$. But: $\xi_j\mu_{X_j}=\xi_j\T(\xi_j)$ holds by definition of an algebra structure map $\xi_j$, so indeed the above fork commutes.
Thus $g=c\circ p$ for a unique homomorphism $c:(X,\xi)\to(\sigma,\varsigma)$. We can compute: $$\begin{align}c\circ h_j&=c\circ p\circ\eta_{\sum_i X_i}\circ j\\&=g\circ\T(j)\circ\eta_{X_j}\\&=c_j\circ\xi_j\circ\eta_{X_j}\\&=c_j\end{align}$$As desired. Moreover, if $\zeta:(X,\xi)\to(\sigma,\varsigma)$ is any other homomorphism with the property $\zeta\circ h_j=c_j$ for all $j\in I$, then: $$\zeta\circ p\circ\eta_{\sum_i X_i}\circ j\equiv g\circ\eta_{\sum_i X_i}\circ j,\,\forall j\in I$$Implies by the uniqueness of the coproduct arrow that: $$\zeta\circ p\circ\eta_{\sum_i X_i}=g\circ\eta_{\sum_i X_i}$$Then we use the fact that $\zeta,g$ and $p$ are all homomorphisms to get: $$\begin{align}g&=g\circ\mu_{\sum_i X_i}\circ\T(\eta_{\sum_i X_i})\\&=\varsigma\circ\T(g\circ\eta_{\sum_i X_i})\\&=\varsigma\circ\T(\zeta)\circ \T(p)\circ\T(\eta_{\sum_i X_i})\\&=\zeta\circ\xi\circ\T(p)\circ\T(\eta_{\sum_i X_i})\\&=\zeta\circ p\circ\mu_{\sum_i X_i}\circ\T(\eta_{\sum_i X_i})\\&=\zeta\circ p\end{align}$$But $g=\zeta\circ p$ implies $\zeta=c$ by uniqueness! Therefore, $c$ is the unique homomorphism with the property $c\circ h_j=c_j$ for all $j$.
That concludes $(X,\xi)$ is indeed a coproduct for $(X_i,\xi_i)_{i\in I}$.
A: One approach to such problems is to establish more broadly a natural bijection of cocones with vertex $(A,\alpha)$ under the diagram consisting of the parallel morphisms $T\bigsqcup_i\alpha_i$ and $\mu_{\bigsqcup_i A_i}\circ T\kappa$, and cocones with vertex $(A,\alpha)$ under the discrete diagram of objects $(A_i,\alpha_i)$. Given such a natural bijection, the colimits of the diagrams would have to correspond, whence the coequalizer and coproduct would. In particular, such a natural bijection would explicitly describe how to go back and forth between components of the cocone under the parallel morphisms and components of the cocone under the discrete diagram.

A cocone with vertex $(A,\alpha)$ under $(A_i,\alpha_i)$ in the category of $T$-algebras is given by a cocone with vertex $A$ and components $f_i\colon A\to A_i$ that are $T$-algebra homomorphisms, i.e. satisfying $f_i\circ\alpha_i=\alpha\circ f_i$.
Let $j_i\colon A_i\to\bigsqcup_i A_i$ be inclusions of a coproduct, i.e.a universal cocone. Then by definition cocones $f_i\colon A_i\to A$ correspond bijectively to morphisms $f\colon\bigsqcup_i A_i\to A$ such that $f\circ j_i=f_i$. Moreover, the universal property of the unit $\eta_{\bigsqcup_i A_i}\to T\bigsqcup_i A_i$ asserts that morphisms $f\colon\bigsqcup_i A_i\to A$ correspond bijectively to morphisms of $T$-algebras $\phi\colon T\bigsqcup_i A_i\to A$ from $(T\bigsqcup_i A_i,\mu_{T\bigsqcup_i A_i})$ to $(A,\alpha)$ such that $\phi\circ\eta_{\bigsqcup_i A_i}=f$.
The condition $f_i\circ\alpha_i=\alpha\circ Tf_i$ that the cocone $f_i\colon A_i\to A$ consits of $T$-algebra homomorphisms from $(A_i,\alpha_i)$ to $(A,\alpha)$ can be expressed in terms of the corresponding $T$-algebra homomorphisms $\phi\colon T\bigsqcup_iA_i\to A$ as $\phi\circ\eta_{\bigsqcup_iA_i}\circ j_i\circ\alpha_i=\alpha\circ T\phi\circ T\eta_{\bigsqcup_iA_i}\circ Tj_i$.
To simplify the conditions, let $k_i\colon TA_i\to\bigsqcup_i TA_i$ be coproduct inclusions. Then by definition $\bigsqcup_i\alpha_i\colon\bigsqcup_iTA_i\to\bigsqcup_iA_i$ satisfies $j_i\circ\alpha_i=\bigsqcup_i\alpha_i\circ k_i$. Moreover, $\eta_{\bigsqcup_i A_i}\circ\bigsqcup_i\alpha_i=T\bigsqcup_i\alpha_i\circ\eta_{\bigsqcup_i TA_i}$ by naturality of the unit of the monad. Thus the left-hand side simplifies to $\phi\circ\bigsqcup_i\alpha_i\circ\eta_{\bigsqcup_iTA_i}\circ k_i$.
Next, $\eta_{\bigsqcup_i A_i}\circ j_i=Tj_i\circ\eta_{A_i}$ by the naturality of the unit. Moreover, $Tj_i\colon TA_i\to T\bigsqcup_iA_i$ factor as $Tj_i=\kappa\circ k_i$ for a unique $\kappa\colon\bigsqcup_i TA_i\to T\bigsqcup_iA_i$. Since $\phi\colon T\bigsqcup_iA_i\to A$ is a $T$-algebra homomorphism from $(T\bigsqcup_iA_i,\mu_{\bigsqcup_iA_i})$ to $(A,\alpha)$, we also have $\alpha\circ T\phi=\phi\circ\mu_{\bigsqcup_iA_i}$.
Finally, the unit of the monad being a unit for its multiplication gives $\mu_{\bigsqcup_iA_i}\circ T\eta_{\bigsqcup_iA_i}=\mu_{\bigsqcup_iA_i}\circ\eta_{T\bigsqcup_iA_i}$, and naturality of the unit then yields $\eta_{T\bigsqcup_iA_i}\circ\kappa=T\kappa\circ\eta_{\bigsqcup_iA_i}$. Thus the right-hand side simplifies to $\phi\circ\mu_{\bigsqcup_iA_i}\circ T\eta_{\bigsqcup_iA_i}\circ\kappa\circ k_i$.
We now have that cocones $f_i\colon A_i\to A$ of $T$-algebra homomorphisms correspond naturally to $T$-algbera homorphisms $\phi\colon T\bigsqcup_i A_i\to A$ satisfying $\phi\circ\bigsqcup_i\alpha_i\circ\eta_{\bigsqcup_iTA_i}\circ k_i=\phi\circ\mu_{\bigsqcup_iA_i}\circ T\kappa\circ\eta_{\bigsqcup_iTA_i}\circ k_i$ for all $i$.
Since $k_i\colon TA_i\to\bigsqcup_iTA_i$ is a univeral cocone, this is equivalent to $\phi\circ\bigsqcup_i\alpha_i\circ\eta_{\bigsqcup_iTA_i}=\phi\circ\mu_{\bigsqcup_iA_i}\circ T\kappa\circ\eta_{\bigsqcup_iTA_i}$.
Moreover, the universal property of the unit $\eta_{\bigsqcup_iA_i}\colon\bigsqcup_i A_i\to T\bigsqcup_i A_i$ implies the condition is equivalent to the equation $\phi\circ\bigsqcup_i\alpha_i\circ=\phi\circ\mu_{\bigsqcup_iA_i}\circ T\kappa$ in the category of $T$-algebra homomorphisms. This is then the desired natural bijection between cocones under the discrete diagram and cocones under the parallel pair of morphisms.
