How can I define this natural transformation $X^{A+B} \to X^A \times X^B$?

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So, I have to define a natural transformation between functors $F,G:C\to C$, where $F(X)= X^{A+B}$ and $G(X)=X^A\times X^B$. I figured out that $\alpha_X:X^{A+B}\to X^A\times X^B$ must be $\langle f,g \rangle$ where $f:X^{A+B}\to X^A$ and $g:X^{A+B}\to X^B$, but I can't define $f$ (or $g$, it's the same). How can I do it?

Thanks in advance.

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By definition of the exponential object, $\hom(X^{A+B}, X^A) \cong \hom(X^{A+B} \times A, X)$. So you want a natural morphism $X^{A+B} \times A \to X$ (and similarly for $B$, then you take the product of these two morphisms).

But again by definition of the exponential object you have a natural morphism: $$\operatorname{eval} : X^{A+B} \times (A+B) \to X.$$

By the universal properties of the product and the coproduct, $$X^{A+B} \times (A + B) \cong (X^{A+B} \times A) + (X^{A+B} \times B).$$

So if you compose the evaluation with the canonical (natural) "injection": $$X^{A+B} \times A \to (X^{A+B} \times A) + (X^{A+B} \times B),$$ you get the morphism you want. I will let you check that everything is natural (a healthy exercise).

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The exponential object is contravariant in the "exponent" variable: If $A \to B$ is a morphism, this induces a morphism of functors $$\hom(-,X^B)\cong \hom(- \times B,X) \to \hom(- \times A,X) \cong \hom(-,X^A) $$ and hence a morphism $X^B \to X^A$ (Yoneda Lemma). This morphism is natural in $X$ since the morphisms of functors above are natural in $X$.

Thus, the inclusions $A \to A+B \leftarrow B$ induce morphisms $X^A \leftarrow X^{A \times B} \rightarrow X^B$. By the universal property of the product these yield a morphism $X^{A + B} \to X^A \times X^B$. This morphism is natural in $X$ since its components are.

Actually this morphism is an isomorphism when we have the "distributive law" $A \times C + B \times C \cong (A+B) \times C$ (via the canonical morphism $\to$) in our category. This can be seen best via the Yoneda Lemma: Just write down the represented functors and observe that they are isomorphic.