The forgetful functor $U:T-Alg\rightarrow C$ preserves limits

Question

Suppose $F$ is an endofunctor of a complete category $C$.

Let $F-Alg$ be the category that has objects the pairs $(X, \ a:FX\rightarrow X)$ where $X$ is an object in $C$, and that has morphisms $f:(X,a)\rightarrow (Y,b)$ where $f$ is a morphisms in $C$ with the property $fa=b\circ Ff$.

Let $M:I\rightarrow F-Alg$ be a diagram with limit $((\lim M,m),\mu_i)$. Show that the forgetful functor $U:F-Alg\rightarrow C$ preserves this limit.

My attempt.

I need to show that the limit of $UM$ in $C$ is $(\lim M,\mu_i)$. So suppose $X$ is an object together with maps $\nu_i:X\rightarrow object(Mi)$ satisfying $M\alpha\circ \nu_i=\nu_j$ for each $\alpha:i\rightarrow j$ in $I$. My job is now to find a map $s:X\rightarrow \lim M$ in $C$ with the property $\mu_i\circ s= \nu_i$.

Of course, by the universal property of limits in $F-Alg$, whenever we have an object $(X,a)$ together with morphisms $\nu_i:(X,a)\rightarrow Mi$ satisfying $M\alpha\circ \nu_i=\nu_j$, there is a unique $k:(X,a)\rightarrow (\lim M,m)$ with the property $\mu_i\circ k= \nu_i$. I would like to choose $s=k$. The issue I am having is that we don’t know if the object $X$ in $C$ has a corresponding object $(X,a)$ in $F-Alg$, because it is not guaranteed that a map $a:FX\rightarrow X$ exists in $C$. I tried to raise the map $k$ to $Fk$ and use it is a morphism in $F-Alg$, but with no success...

How do I ensure $X$ comes with a morphism $a:FX\rightarrow X$?

I added a sketch for clarity



I would prefer a hint over an answer.

Answer 1

Hint: For simplicity, I'll consider the case of a product of two objects. Working backwards from the conclusion, if you have two $T$-algebras $(X, f)$ and $(Y, g)$, then the product of these two $T$-algebras will end up being isomorphic to $(X \times Y, h)$ for some $h$. So, a good place to start would probably be to look for an explicit way to construct a morphism $h : T(X\times Y) \to X\times Y$. Then, if you can show that this $T$-algebra is a product of $(X, f)$ and $(Y, g)$, you will be done for this type of limit by the uniqueness of limits. (And then, hopefully, your argument will be straightforward to generalize to more general limits.)