Let $\mathcal{C}$ be a small category and let $\hat{\mathcal{C}}$ be its category of presheaves. I want to show that every representable presheaf $y_C\in \hat{\mathcal{C}}$ (for some $C\in\mathcal{C}$) is projective. $y$ here is the Yoneda embedding. So suppose that $F,G\in \hat{\mathcal{C}}$ also, with natural transformations $f:y_C\to F$ and $g:G\to F$, with $g$ epi. Then using the axiom of choice, it is rather easy to see that for each $D\in\mathcal{C}$, there exists a function $h_D: y_C D\to GD$ such that $g_D h_D= f_D.$ If $h=(h_D)_{D\in\mathcal{C}}$ is a natural transformation ($y_C\to G$) then it is clear that $gh=f.$ My problem now is with showing that $h$ is indeed a natural transformation.
So for any map $k:D\to D'$ in $\mathcal{C}^{op}$, we need that $(Gk)(h_D) = h_{D'}(y_Ck).$ This is where I am stuck. It is not clear to my why this should hold. We know what $y_C k$ is, but not what $Gk$ behaves like. I tried involving $FD$ and $FD'$ in the diagram, but this has not helped me.
I want to show that the outer square commutes:
It is clear that the upper and lower inner squares commute since $f$ and $g$ are natural transformations, and it is clear that the left and right triangles commute by definition of $h_D$ and $h_{D'}$, but that alone does not seem to be enough for the outer square to commute.
How can I complete the proof? Any hints would be appreciated. I think I should involve the definition of $y_C$ somehow, since I have not actually used that yet.