Does the usual development of category theory (within Goedel-Bernays set theory, for example) require the axiom of replacement? I would have asserted that this was obviously true, but it seems to be common wisdom that the axiom of replacement is an exotic axiom not used outside of axiomatic set theory. Also, sadly the overlap between things that I have thought were obviously true and were in fact false is larger than I would like to admit...
The axiom of replacement basically says that if a class is the same size as a set, then it is a set. This allows us to identify classes that are sets as being small and those that are proper classes as large. Without replacement, you could have countable classes that not sets.
I don't see how to construct most limits and colimits in familiar categories such as Set or Top without the use of replacement. Already, for an infinite set $X$, I don't see how you construct $$ \coprod_{i=0}^\infty P^i X, $$ where $P^i$ is the $i$-th power set of $X$ (i.e. $P^0 X = X$ and $P^i X = P (P^{i-1}X$)).
Using replacement, it's easy to construct. You form the set $$ \bigcup_{i=0}^\infty P^i X, $$ and then the coproduct is isomorphic to the set of pairs $(i, x)$, where $x \in P^i X$. Am I completely misunderstanding the issue here, and this coproduct can be proven to exist without replacement?
I suppose you could cook up a definition of diagram so that in the absence of replacement, you cannot even form the diagram for this particular coproduct, but that would be sort-of unsatisfying.
$\bigcup_n P^n(X)$
in the absence of replacement is not the coproduct but the iteration of the power set. It is provable in Zermelo set theory (ZF without replacement) that any set-indexed family of sets has a coproduct that is a set. (Limits and colimits of small-indexed diagrams of sets are OK too.) But Zermelo set theory does not prove the existence of the sequence of iterated power sets$n\mapsto P^n(X)$
when $X$ is, for example, the set of natural numbers. $\endgroup$