Prove that $a$ and $b$ have a coproduct in $C$ iff $C(a, -) \times C(b, -)$ is representable From Mac Lane's Category theory:

Prove that $a$ and $b$ have a coproduct in $C$ iff $C(a, -) \times C(b, -)$ is representable.

I can see that if $a,b$ have a coproduct $a \coprod b$, then the representation is $\psi : C(a, -) \times C(b,-)\rightarrow C(a \coprod b, -)$ with $\psi : \langle f=u \circ i,g = u \circ j \rangle \mapsto u$ where $u$ is the unique map in the coproduct diagram.
But I'm having trouble proving the reverse direction.  I can't see exactly how I'd find an $i: a \rightarrow r$ and $j : b \rightarrow r$ such that the coproduct requirement holds.
Anyone have any ideas?
 A: Derek Elkins gave you what you need. Suppose $r$ is the representing object, so there is a natural equivalence of functors $\eta: C(a,-) \times C(b,-) \to C(r,-)$.  The point, of course, is that $r$ serves as a coproduct of $a$ and $b$.  So, you need only check that $r$ satisfies the right universal mapping property.  For this, you need the coproduct "inclusion" maps.  Evaluate $\eta$ at $r$ so that we have a bijection
$$
\eta_r:  C(a,r) \times C(b,r) \to C(r,r).
$$
Hence  we have $\eta_r(\alpha, \beta)=1_r$ for a unique pair of maps $\alpha: a \to r$ and $\beta : b \to r$.  Now, check that the setup $(r, \alpha, \beta)$ meets all the properties of the coproduct $a \amalg b$.  It does, and for this you will need the naturality of $\eta$. 
Addendum:  Take all the setup and notation above.  Suppose we are given two arbitrary maps $u: a \to d$ and $v: b \to d$ for some object $d$.  We must show there is a unique $f: r \to d$ such that $f \circ \alpha =u$ and $f \circ \beta = v$.  Evaluate $\eta$ at $d$ (honestly, the only thing you could do here) to get the bijection 
$$
\eta_d: C(a,d) \times C(b,d) \to C(r,d).
$$
Put $f=\eta_d(u,v)$.  This solves the commutativity problems by naturality:
$$
(f_* \circ \eta_r)(\alpha, \beta) = f_*(1_r)= f
$$
around two legs of the square while the other two legs give
$$
(\eta_d \circ (f_* \times f_*))(\alpha, \beta) = \eta_d (f\circ \alpha, f\circ \beta).
$$
But by naturality this has to be $f$, so 
$$
\eta_d (f\circ \alpha, f\circ \beta) =  f = \eta_d(u,v).
$$
But $\eta_d$ is one-to-one, so $f \circ \alpha = u$ and $f \circ \beta =v$. To show that $f$ is unique.... (you should handle this)
