Categorical objects defined by both limits and colimits

Question

I'm relearning category theory after a while, and I have a somewhat silly question.

When I think about whether a particular structure of interest is a limit or colimit of a particular category, I think about whether there is a natural projection (for limits) or embedding (for colimits) of the limit/colimit in/out of the limiting objects.

However, in cases where there are natural embeddings and projections, is it possible to define the structure both ways?

Here's a concrete example. Consider the category of pointed sets (sets with a distinguished element, with morphisms as set-theoretic functions mapping distinguished elements to other distinguished elements). Obviously we could define the cartesian product on this category via a limit the same way we would in Set. But, there is a natural embedding map too. $f:A_i\mapsto\prod\limits_{i\in I}A_i$ can be defined by $f(x)=(a_j)_{j\in I}$, with $a_j=x$ when $j=i$, else, $a_j=p_j$, where $p_j$ is the distinguished element of $A_j$.

Do these embeddings lend themselves to a colimit definition of the cartesian product on this category?

Answer 1

The product of pointed sets is not a colimit in any reasonable way. What you described holds more generally in any category with zero morphisms. If $(A_i)_{i \in I}$ is a family whose product exists, then there is a split monomorphism $$A_j \hookrightarrow \prod_{i \in I} A_i$$ for every index $j$, namely by applying the universal property to $0 : A_j \to A_i$ for $i \neq j$ and $\mathrm{id} : A_j \to A_j$. The splitting is the $j$th projection. In an additive category, for finite sets $I$ these monomorphisms form a coproduct cocone, so that $\prod_{i \in I} A_i \cong \coprod_{i \in I} A_i$. But when $I$ is infinite, maps on a product tend to me hard to describe, and sometimes $\prod_{i \in I} A_i$ turns out to be some sort of "completion" of the coproduct.

There are some situations where pullback squares are also pushout squares. For example, when $G$ is an abelian group with two subgroups $U,V$, then

$$\begin{array}{ccc} U \cap V & \rightarrow & U \\ \downarrow && \downarrow \\ V & \rightarrow & U+V \end{array}$$ is a pullback square of abelian groups for trivial reasons, but interestingly also a pushout square.

Answer 2

$\require{AMScd}$A more elaborate, although not so different, example of a pullback that's also a pushout comes from homological algebra: let $$ \begin{CD} A @>u>> B @>v>> C\end{CD}$$ be a complex of morphisms in (say) abelian groups; this means $vu=0$, i.e. $I=\text{im }u\subseteq \ker v = K$, so $H(u,v):=K/I$ is defined.

Then the square $$ \begin{CD} K @>p>> H(u,v) \\ @ViVV @VVjV \\ A @>>q> A/I \end{CD} $$ where $i$ is the embedding and $p,q$ are projections to the quotient, is both a pullback and a pushout.