Assume $\mathcal{C}$ and $\mathcal{D}$ are two categories where finite products exist, then I have proved that if a functor $F: \mathcal{C}\rightarrow \mathcal{D}$ preserves finite products, then we have $F$ preserves equalizer if and only if $F$ preserves pullback diagram.
Then I want to ask if this holds without the condition that $F$ preserves finite products. If the answer is no, I want to know the counterexamples for the “if” part and the "only if " part respectively.
Thanks!
Consider a partially ordered set as a category $\mathcal{C}$. Since there is at most one morphism between any two objects in $\mathcal{C}$, equalisers exist but are always trivial, and any functor out of $\mathcal{C}$ will preserve equalisers. On the other hand, pullbacks may or may not exist in $\mathcal{C}$, and may or may not be preserved by functors out of $\mathcal{C}$. Therefore a functor that preserves equalisers may not preserve pullbacks.
It is more difficult to construct a functor that preserves pullbacks but not equalisers, because one can construct equalisers using pullbacks and "very weak" coequalisers. More precisely, given a parallel pair $X \rightrightarrows Y$, if there is a morphism $Y \to Z$ such that the two composites $X \rightrightarrows Y \to Z$ are equal – no universality, not even weak universality, required – then we can construct the equaliser of $X \rightrightarrows Y$ as the pullback of $Y \to Y \times_Z Y$ along $X \to Y \times_Z Y$. In a category $\mathcal{D}$ where every parallel pair has a "very weak" coequaliser, such as a category with a terminal object, any functor out of $\mathcal{D}$ that preserves pullbacks will automatically preserve equalisers as well. (Any functor whatsoever preserves "very weak" coequalisers.)
There should be a counterexample but it seems to be more fiddly than I first imagined. Perhaps I will add one if I figure out an easy one.