Extending limit existence theorem to functors

If a functor $G:\mathcal{C}\rightarrow \mathcal{D}$ preserves all products and equalizers on pairs, is it continuous?

It seems like we can apply the same construction as the construction of arbitrary limits from products and equalizers in categories. In Mac Lane's Categories for the Working Mathematician Chapter 5 Section 4 Exercise 2, we seem to be given an extra assumption that $\mathcal{C}$ is complete, but I don't see why this is necessary.

• What is that exercise? – Berci Mar 25 '17 at 22:51
• @Berci paraphrased, the exercises says: If $\mathcal{C}$ is complete, and $G:\mathcal{C}\rightarrow \mathcal{D}$ preserves products and equalizers, prove $G$ is continuous. – VF1 Mar 25 '17 at 23:35
• If $\mathcal{C}$ is not complete, one has to wonder about limits in $\mathcal{C}$ that cannot be written as an equalizer of a product. – Hurkyl Mar 26 '17 at 5:02
• The statement is true if $\mathcal{C}$ has all products (because then each limit has the usual description as an equalizer of morphisms between products). In general it is false. (Actually, many people seem to believe that it is true, because they omit the assumption that $\mathcal{C}$ has products.) Perhaps I can find the counterexample ... – HeinrichD Mar 26 '17 at 10:11
• It was discussed here two years ago, and a counterexample was mentioned (just dualize it): math.stackexchange.com/questions/1198452 ... so perhaps this question can be closed as a duplicate. – HeinrichD Mar 26 '17 at 10:25