Enough projectives and $F$ preserves limits implies $G$ preserves epi's. Exercise: Let $\mathcal{C}, \mathcal{D}$ be categories, $G : \mathcal{C} \to \mathcal{D}$ and $F : \mathcal{D} \to \mathcal{C}$ an adjunction $F \dashv G$. Suppose $\mathcal{D}$ has enough projectives and $F$ preserves projectives. Prove that $G$ preserves epimorphisms.
My try: Suppose $f : x \to y \in \mathcal{C}$ is epi, then we have to show that $Gf : Gx \to Gy$ is epi. That is, for all $g, h : Gy \to z$ in $\mathcal{D}$ we must have $gFf = hFf$ implies $g = h$. I have tried writing down an projective for $Fx$ and $Fy$ but that seems to lead nowhere.

Suppose $f : X \to Y$ is an epimorphism in $\mathcal{C}$. We wish to show that $G f : G X \to G Y$ is an epimorphism. 
Let $q : B \to G Y$ be an epimorphism in $\mathcal{D}$ where $B$ is projective. Then $F B$ is projective, so we have a morphism $x : F B \to X$ in $\mathcal{C}$ such that $f \circ x = \epsilon_Y \circ F q$, where $\epsilon_Y : F G Y \to Y$ is the adjunction counit. Hence, $G f \circ G x \circ \eta_B = q$. But $q : B \to G Y$ is an epimorphism, so $G f : G X \to G Y$ is also an epimorphism.
