Unique extension of a map from $X$ to a map from the free algebra

Question

Suppose we are in a category $\mathcal{C}$ and we have monad on this category, written as $(P,\sigma, \mu)$. We can consider the category of $P$-algebras. Among its elements is are the free algebras, which are of the form $(P(X),\mu_X)$, for some object $X$. In most cases, like with groups or monoids, the construction of such a 'free' object has the following universal property: If there is another object $T$, and a map $t:X\to T$, then this extends in a unique way to a lift $\hat{t}:P(X)\to T$, making the obvious triangle commute. It seems to me that this also holds in the more abstract case, for an arbitrary category and an arbitrary monad. However, I fail to find a source which elaborates on this subject explicitly, and only found a source which just mentions it as a result. Does anyone have an argument why this should, or should not, be true?

Thanks in advance!

Answer 1

In the comments, you say you are happy with the adjunction $F^P \dashv U^P$, where $U^P\colon \mathsf{P{-}Alg}\to C$ takes a $P$-algebra $(X,\alpha)$ to its underlying object $X$ and $F^P\colon C\to \mathsf{P{-}Alg}$ takes an object $X$ to the free algebra $(P(X),\mu_X)$. This adjunction means exactly the same thing as the universal property you're interested in - it's just a more concise way of writing it down.

Let $X$ be an object in $C$ and let $(Y,\alpha)$ be a $P$-algebra. Then the adjunction tells us that $\text{Hom}_{\mathcal{C}}(X,Y) = \text{Hom}_{\mathcal{C}}(X,U^P(Y,\alpha)) \cong \text{Hom}_{\mathsf{P{-}Alg}}(F^P(X),(Y,\alpha))$. So for every arrow in $C$ from $X$ to the underlying object of the $P$-algebra $(Y,\alpha)$, there is a unique $P$-algebra morphism $F^P(X)\to (Y,\alpha)$ from the free $P$-algebra on $X$ to $(Y,\alpha)$.

Now you'd like to know something a bit more - not just that there is a natural bijection between $\text{Hom}_{\mathcal{C}}(X,Y)$ and $\text{Hom}_{\mathsf{P{-}Alg}}(F^P(X),(Y,\alpha))$, but that the arrow $U^P(f)\colon P(X)\to Y$ defining the $P$-algebra homomorphism $f\colon F^P(X)\to (Y,\alpha)$ "lifts" the corresponding arrow $g\colon X\to Y$ in $C$, in the sense that $U^P(f)\circ\sigma_X = g$. To do this, we need to dig in a bit more into how the adjunction works. And there's a standard way to understand an adjunction as a universal property. You can skip this whole section if you're already familiar with it.

Suppose $F\dashv G$ are adjoint functors, where $F\colon C\to D$ and $G\colon D\to C$. Let $X\in C$. Then in particular, we have a natural bijection $\text{Hom}_D(F(X),F(X))\cong \text{Hom}_C(X,G(F(X)))$. So the identity map $\text{id}_{F(X)}\in \text{Hom}_D(F(X),F(X))$ corresponds to some arrow $\eta_X\in \text{Hom}_C(X,G(F(X)))$. The arrows $\eta_X$ cohere to a natural transformation $\eta\colon \text{Id}_C\to G\circ F$, which is called the unit of the adjunction. Further, naturality tells us that for any arrow $f\colon F(X)\to Y$ in $D$, we have a commutative square: $$\require{AMScd}\begin{CD} \text{Hom}_D(F(X),F(X)) @>{f\circ-}>> \text{Hom}_D(F(X),Y)\\ @V{\cong}VV @V{\cong}VV \\ \text{Hom}_C(X,G(F(X))) @>{G(f)\circ-}>> \text{Hom}_C(X,G(Y)) \end{CD}$$ If we start in the upper left with $\text{id}_{F(X)}$ and go right, we get $f\circ \text{id}_{F(X)} = f$, and if we go down, we get the arrow $X\to G(Y)$ in $C$ corresponding to $f$ in the adjunction. Instead, if we go down first we get $\eta_X$, and then if we go right, we get $G(f)\circ \eta_X$. In other words, $\eta_X$ is a magical arrow with the property that if we want to compute what arrow $X\to G(Y)$ in $C$ corresponds to $f\colon F(X)\to Y$ in $D$, we just have to look at $G(f)\circ \eta_X$. A slogan is that "$\eta_X$ is the universal way to go from $X$ to something in the image of the functor $G$", in the sense that for any other way to go from $X$ to something in the image of $G$, $g\colon X\to G(Y)$, it factors through $\eta_X$ as $g = G(f)\circ \eta_X$, where $f\colon F(X)\to Y$ corresponds to $g$ through the adjunction.

Ok, back to the concrete case of $P$-algebras. If you understand the adjunction $F^P\dashv U^P$, you can compute that for any $X\in C$, $U^P(F^P(X)) = P(X)$, and the unit $\eta_X$ of the adjunction is just the unit of the monad $\sigma_X\colon X\to P(X)$. So if $g\colon X\colon U^P(Y,\alpha) = Y$ is an arrow from $X$ to the underlying object of a $P$-algebra $(Y,\alpha)$, and $f\colon F^P(X)\to (Y,\alpha)$ is the corresponding homomorphism of $P$-algebras, then $f$ and $g$ are related by $g = U^P(f)\circ \eta_X = U^P(f)\circ \sigma_X$. This is exactly the "lifting" commutative triangle we wanted.