Peng Zhou

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====== 2025-01-09  ======
  * colimit of algebras


===== colimit =====
Let's be really naive and simple. What is a colimit? Suppose you have a category $\mathcal{C}$, many objects, morphisms. 
And then, there is a push-out diagram $A \gets B \to C$. This gang of objects talks with everyone, for example, someone called $X$. $A$ talks with $X$, get a set $Hom(A,X)$, so does $B$, get $Hom(B,X)$. It might be big or small, we don't know, depends on $X$'s relation with all of them. Oh, and don't forget $Hom(C,X)$. So, how do we deal with the three sets? We cannot just union them together, that would be stupid. Well, by pre-composition, we have maps $Hom(A,X) \to Hom(C,X) \gets Hom(B,X)$. So, we can ask for compatible arrows in $Hom(A,X)$ and $Hom(B,X)$, see if their image in $Hom(C,X)$ coincide or not. If they coincide, then we remember that. Suppose $M$ is the would-be colimit, then we have $Hom(M,X)$ be the fibre-product. So, we find $M$ by its relations with everybody, that is $M \in Fun(\mathcal{C}, Set)$. It probably is representable. I am not sure what condition do I need to get an actual $M$, instead of indirect evidence. 

OK now, here comes an algebra. $A$, and we have two subalgebra, with some intersections. For example, 
$A=\C[x,y]$, $A_1=\C[x], A_2=\C[y]$, $A_{12} = A_1 \cap A_2 =\C$. So, what is 
$$ colim(\C[y] \gets \C \to \C[x]) $$
No, this is really bad. If you just give me these two subalgebra, I don't know how they will talk. The worst case, they don't talk at all, so you need the free-algebra generated by these guys. And that surely can map to $\C[x,y]$, $\C[x,y]/(xy)$ etc. 

So, we cannot just say, here are some nice looking subalgebra that contains all the generators, then you go off.