Prove that a functor creates limits

Question

Let $F:\textbf{C}\to\textbf{D}$ be a functor. How do we prove that $F$ creates limits for a particular class of diagrams if $\textbf{C}$ has those limits and $F$ prserves them and $F$ reflects isomorphisms?

Answer 1

It's enough to prove that $F$ reflects limits, ie. that if $π = (π_i : C → C_i)_i$ is a cone in $\mathbf C$, and $Fπ$ is the limiting cone, then so is $π$ itself. To do this, take the limit $(λ_i : L → C_i)_i$ of the diagram in question and factor $π$ through it as $π = f ∘ λ$ (ie. $π_i = f ∘ λ_i$ for each $i$). You can't prove directly that $f$ is iso, but since $F$ is conservative, you can just show that $Ff$ is, and that's easy. $Fπ$ is the limiting cone by assumption, $Fλ$ is too because $F$ preserves limits, and $Ff$ has to be the unique isomorphism $g: FC → FL$ which satisfies $Fπ = g ∘ Fλ$.

(This is assuming that you're using the definition of created limit such as the one on nlab. If you use a stronger definition, such as the one on Wikipedia with the uniqueness condition, then this isn't true as stated. For a trivial counterexample you can take the (equivalence) functor $F : 2_\mathrm{ind} → 1$ from the two object indiscrete category to the one object one, and note that the unary product isn't uniquely lifted.)