Products, Naturality and Functors i have the following problem: i want to show the following: Let $\bf{D}$ be a category with binary products, then $\bf{D}^{\bf{C}}$ also has binary products. Remark: $\bf{D}^{\bf{C}}$ is the category with objects functors from $\bf{C}$ to $\bf{D}$ and arrows natural transformations.
My problem is not to define the natural transformations $\pi_1$ and $\pi_2$ from $F\times G$ to $F$ and $G$ resp. (since we have the projections in $\bf{D}$. But now let $h:D\rightarrow F$ and $k:D\rightarrow G$ be two arrows in $\bf{D}^{\bf{C}}$. My question is how to define the unique arrow $u:D\rightarrow F\times G$.
Can someone help me? I am happy about hints, solutions, etc.
Thanks.
 A: I'm assuming that you have understood that (up to natural isomorphism) the functor $F \times G \colon \mathbf C \to \mathbf D$ is just the functor sending every $c \in \mathbf C$ in $F(c) \times G(c)$ and every morphism $f \in \mathbf C(c,c')$ in $F(f)\times G(f)$, that is the unique morphism making the following diagram commute
$$\require{AMScd}\begin{CD} F(c)@<<\pi_{F(c)}< F(c) \times G(c) @>\pi_{G(c)}>>G(c)\\ 
@VF(f)VV @VF(f)\times G(f)VV  @VVG(f)V \\
F(c') @<<\pi_{F(c')}< F(c') \times G(c') @>>\pi_{G(c')}> G(c') \end{CD}$$
If $D \colon \mathbf C \to \mathbf D$ was another functor with natural transformations $h \colon D \to F$ and $k \colon D \to G$ then such natural transformations have components $h_c \colon D(c) \to F(c)$ and $k_c \colon D(c) \to G(c)$ for every $c \in \mathbf C$. 
By universal property of the product then there's a unique morphism $(h,k)_c \colon D(c) \to F(c) \times G(c)$ such that $\pi_{F(c)} \circ (h,k)_c=h_c$ and $\pi_{G(c)} \circ (h,k)_c=k_c$.
It's now a diagram chase to prove that the morphisms $(h,k)_c$ are natural and give the only natural transformation from $D$ to $F \times G$ such that $\pi_F \circ (h,k) = h$ and $\pi_G \circ (h,k)=k$: where by $\pi_F$ and $\pi_G$ I denote the projections of the functor-product.
