Proof of $C^{A+B} \cong C^{A} \times C^{B}$ using only UMP and definitions of products and exponential objects I need to show that in any category with binary products and coproducts, the following holds:
$C^{A+B} \cong C^{A} \times C^{B}$
using only universal mapping properties and definitions of (co)products and exponential objects.
I'm aware of this question:
At a closed monoidal category, how can I derive a morphism $C^A\times C^B\to C^{A+B}$?,
but I'd like to look at a proof which does not involve the Yoneda lemma, right adjoints, binatural bijections etc.
As the problem statement
http://www.andrew.cmu.edu/user/awodey/SummerSchool/HW/cathw2.pdf 
tells us, it can be shown by "drawing commutative diagrams" only:

There is no trick for this one -- you'll have to crank it out


Actually, this is what I was asking for at first place, at the question you're linking. But then I saw it gets really messy without the Yoneda lemma.
My (incomplete) work on this was the following: You have to find isomorphic arrows $f:C^{A+B}\to C^A\times C^B$ and $g:C^A\times C^B\to C^{A+B}$ such, that their composition gives the corresponding identity arrow.

*

*With $\pi_i:A_1\times A_2\to A_i$ I denote the projections of product $A_1\times A_2$.

*With $\kappa_i^{A_1,A_2}:A_i\to A_1+A_2$ I denote the coprojections of coproduct $A_1+A_2$.

*For $f:A\times B\to C$, I denote with $\tilde{f}: A\to C^B$ the currying of $f$ and, for $g:A\to C^B$, I denote with $\bar{g}:A\times B\to C$ its uncurrying.

*$\gamma$ is the braiding iso: $\gamma_{A,B}=\langle\pi_2^{A,B},\pi_1^{A,B}\rangle:A\times B\to B\times A$
Finding $f$
We have $$\mathbf{1}_{C^{A+B}}\times \kappa_1^{A,B}:C^{A+B}\times A\to C^{A+B}\times(A+B)$$ and $$\mathbf{1}_{C^{A+B}}\times \kappa_2^{A,B}:C^{A+B}\times B\to C^{A+B}\times(A+B)$$
Then, put $h_i$ to be:
$$h_1=\varepsilon\circ(\mathbf{1}_{C^{A+B}}\times \kappa_1^{A,B}):C^{A+B}\times A\to C$$ and $$h_2=\varepsilon\circ(\mathbf{1}_{C^{A+B}}\times \kappa_2^{A,B}):C^{A+B}\times B\to C$$
So, by currying you get $\tilde{h_1}:C^{A+B}\to C^A$ and $\tilde{h_2}:C^{A+B}\to C^B$ and their product will be $$f=\langle\tilde{h_1}, \tilde{h_2}\rangle$$
Finding $g$
You have $$\kappa_1^{(C^A\times C^B)\times A,(C^A\times C^B)\times B}\circ\gamma_{A,C^A\times C^B}:A\times(C^A\times C^B)\to ((C^A\times C^B)\times A)+((C^A\times C^B)\times B)$$ and
$$\kappa_2^{(C^A\times C^B)\times A,(C^A\times C^B)\times B}\circ\gamma_{B,C^A\times C^B}:B\times(C^A\times C^B)\to ((C^A\times C^B)\times A)+((C^A\times C^B)\times B)$$
Then, take their currying and put
$$q_1=\widetilde{\kappa_1^{(C^A\times C^B)\times A,(C^A\times C^B)\times B}\circ\gamma_{A,C^A\times C^B}}:A\to \left(((C^A\times C^B)\times A)+((C^A\times C^B)\times B)\right)^{C^A\times C^B}$$ and $$q_2=\widetilde{\kappa_2^{(C^A\times C^B)\times A,(C^A\times C^B)\times B}\circ\gamma_{B,C^A\times C^B}}:B\to \left(((C^A\times C^B)\times A)+((C^A\times C^B)\times B)\right)^{C^A\times C^B}$$
Their coproduct is $$[q_1,q_2]:A+B\to\left(((C^A\times C^B)\times A)+((C^A\times C^B)\times B)\right)^{C^A\times C^B}$$
Its exponential transpose then is a very helpful arrow (you can check that it is also an iso):
$$t=\overline{[q_1,q_2]}:C^A\times C^B\times (A+B)\to ((C^A\times C^B)\times A)+((C^A\times C^B)\times B)$$
Then, you can easily find the desired arrow $g:C^A\times C^B\to C^{A+B}$ as the exponential transpose of the composition of $t$ with some other arrows including evaluation arrows $\varepsilon_{A,C}$ and $\varepsilon_{B,C}$.
You're not done yet. You have to show that $f\circ g=\mathbf{1}_{C^A\times C^B}$ and $g\circ f=\mathbf{1}_{C^{A+B}}$. At this point, I didn't bother verifying... I quit and just used the Yoneda lemma.
