Previous: [[Coproducts in the category of abelian groups]] After all the discussion with limits, there's an interesting thing we can notice: in most categories, like that of groups, vector spaces, and topological spaces, the categorical product $A \times B$ has as its underlying set the categorical product in $\mathbf{Set}$. However, the coproduct is wildly different. In $\mathbf{Set}$ as in $\mathbf{Top}$ it is the disjoint union $A \sqcup B$, in $\mathbf{Vect}$ it is the direct sum $A \oplus B$, which for finite sums *is* also the product $A \times B$ which is the underlying set. The worst of course is $\mathbf{Grp}$, where the underlying set of the coproduct $A \ast B$ is this awful infinite set of all words from letters in $A$ and $B$. Why is this the case? To understand this, let's think about what's happening "formally" when we look at the underlying set of an object. This is just the forgetful functor $L: \mathcal C \to \mathbf{Set}$ that assigns to an object its set of elements. Then the question we are asking is, for some objects $A$ and $B$ in $\mathcal C$ is it true that $L(A \times B) \cong L(A) \times L(B)$, and similarly for coproducts. To investigate this further, we'll recall the concept of adjoint functors that I mentioned previously in [[The universal property of the free group]]. Two functors $F : \mathcal C \to \mathcal D$ and $G: \mathcal D \to \mathcal C$ are called adjoint functors if for any $A \in \mathcal C$ and $B \in \mathcal D$^[using set membership here is probably the wrong notation] $\operatorname{Hom}_{\mathcal D}(F(A), B)\cong \operatorname{Hom}_{\mathcal C}(A, G(B))$ which is to say that to every morphism between $F(A)$ and $B$ we can associate a morphism between $A$ and $G(B)$. Then $F$ is called "left-adjoint" to $G$ and $G$ is called "right-adjoint" to $F$. The universal property of the free group is exactly that $\operatorname{Hom}_{\mathbf{Set}}(S, \text{Forget}(G)) \cong \operatorname{Hom}_{\mathbf{Grp}}(\text{Free}(S), G)$ This holds in a lot of categories where there is a "free" construction, and is known as the "free-forgetful adjunction." A "free" functor can be *defined* as an adjoint to the forgetful functor! The main theorem that ends up helping us says that "right adjoints preserve limits" and "left adjoints preserve colimits." This says that if we take a limit (resp. colimit) in a category and then apply a functor $F$ that is left-adjoint (resp. right-adjoint) to any other functor, then $F(L)$ where $L$ is the limit (resp. colimit) is the same as the limit (resp. collimit) of the objects under $F$. Then, if we remember that the product is defined as a limit, this means *exactly* that the forgetful functor preserves products, which means that the underlying set of a product is the same as the product of the underlying sets. The coproduct is a *colimit* and the forgetful functor is under no obligation to preserve it. The *free* functor, though, is. And indeed, if $F$ is the free group functor, then $F(S\sqcup T) = F(S)\ast F(T)$ Using this we can analyze coproducts in other categories you might have seen. There is one interesting case, the details of which I'll let you flesh out if you know some point-set topology. In the category $\mathbf{Top}$ of topological spaces and continuous maps, there is an obvious forgetful functor which sends a topological space to its underlying set. However, given a set, what is the obvious "free" topology on it? Is it the discrete topology where every set is open, or the indiscrete topology where no set except the empty set and the whole set is open? As it turns out, *both* are an example of a free construction, and so the forgetful functor has both a left-adjoint and a right-adjoint. This means that the forgetful functor itself is both a right-adjoint and a left-adjoint, so it preserves both limits and colimits. Since (and this is not obvious, but needs proof) both $\mathbf{Set}$ and $\mathbf{Top}$ have all limits and colimits, every (co)limit construction in $\mathbf{Top}$ always has the same underlying set as that construction in $\mathbf{Set}$. Next: [[A worked example - the abelianization functor]]