Is it possible that the inclusion functor does not preserve limits? Let $\mathcal{B}$ be a full subcategory of a category $\mathcal{C}$. Is it possible that the inclusion functor does not preserve the pullbacks (or any limits) that exist in $\mathcal{B}$? I read this somewhere and looks to me like a typo. Is it not? The hom sets are the same, how is such a thing possible, if this is true?
 A: Another example, concrete in a different, informal sense than Hagen von Eitzen's example: Let $\mathcal B$ be the partially ordered set, viewed as a category in the usual way, consisting of three elements, say 0,1,and 2, with $0\leq1$ and $0\leq 2$ while 1 and 2 are incomparable.  Let $\mathcal C$ be the partially ordered set obtained by adjoining to $\mathcal B$ one more element, say 4, with the ordering relations $0\leq 4$, $4\leq1$, and $4\leq 2$.  Then the product (which in partially ordered sets coincides with the greatest lower bound) of 1 and 2 is 0 in $\mathcal B$ but is 4 in $\mathcal C$.  If you want an example using pullbacks rather than products, adjoin a top element 5 to both $\mathcal B$ and $\mathcal C$.
An even simpler example uses the product of an empty family, i.e., a terminal object. Let $\mathcal B$ be the category consisting of just one object and its identity morphism.  Obtain $\mathcal C$ by adjoining a second object and its identity morphism.  Then the object in $\mathcal B$ is terminal there, but $\mathcal C$ has no terminal object.  If you modify $\mathcal C$ by adding a morphism from the object of $\mathcal B$ to the new object, then $\mathcal C$ has a terminal object, but it's not the one from $\mathcal B$.
A: As a concrete example, you may know that the coproduct of groups is the free product and the coproduct of abelian groups is just the direct sum. The free product is never abelian (unless one factor is trivial), hence the inclusion from Ab to Grp certainly fails to preserve coproducts. 
