Isomorphism exists in coproduct of categories The following question was left as an exercise in my class of Algebraic Topology and I am not able to make much progress on it.
Question:  Let $(X_{\alpha})_{\alpha \in A}$ be a family of objects of a category $C$. An object $X$ of $C$ , equipped with morphisms $i_{\alpha} : X_{\alpha} \to X$ is called a coproduct of the $X_{\alpha}$ if it satisfies the following universal property: For all families of morphisms $f_{\alpha} : X_{\alpha} \to Y$ there is a unique $f: X\to Y$ such that $f\circ i_{\alpha} =f_{\alpha}$.

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*Show that if $X$ equipped with the $i_{\alpha}$ is a a coproduct of the $X_{\alpha}$  and if $X'$equipped with the $i'_{\alpha}$ is another coproduct of the $X_{\alpha}$ , then there exists a unique isomorphism $\phi: X \to X'$ such that $\phi \circ i_{\alpha} = i'_{\alpha}$for all $\alpha$.


*Show that coproducts always exist in the following categories: Ens, Top, R-Mod.
In (1) , I am not able to construct such an isomorphism and in (2) , I am aware of the definitions but not able to work. ( Can you please show (2) for one of the categories?)
Do you mind helping me on this problem?
 A: Let $X$ equipped with $i_\alpha$ and $X'$ equipped with $i'_\alpha$ both be coproducts of the $X_\alpha$. In the universal property that makes $X$ equipped with $i_\alpha$ a coproduct, choose $Y=X'$ and $f_\alpha=i'_\alpha$. By the universal property, there is a unique $f:X\to X'$ such that $f\circ i_\alpha=i'_\alpha$. Now we just have to show that this is an isomorphism. Applying the same argument in reverse, we get a $g$ such that $g\circ i'_\alpha=i_\alpha$. Then $(g\circ f)\circ i_\alpha=g\circ(f\circ i_\alpha)=g\circ i'_\alpha=i_\alpha$. But from applying the universal property of $X$ with $Y=X$ and $f_\alpha=i_\alpha$, we know that the morphism $h$ with $h\circ i_\alpha=i_\alpha$ is unique. Since the identity morphism of $X$ satisfies these equations, it follows that $g\circ f$ is that identity morphism. Thus $g$ is the inverse of $f$, so $f$ is an isomorphism.
In the category of topological spaces, the coproduct of $X_\alpha$ is the disjoint union of $X_\alpha$ with the disjoint union topology (whose open sets are the sets whose intersection with each $X_\alpha$ is open), equipped with the injections $i_\alpha$ that map each element to the corresponding element in the disjoint union. These injections are continuous because the disjoint union topology is defined such that it’s the finest topology that makes them continuous; hence the $i_\alpha$ are morphisms.
Given a topological space $Y$ and morphisms (i.e. continuous functions) $f_\alpha:X_\alpha\to Y$, the function $f:X\to Y$ from the disjoint union to $Y$ is defined by sending every element of the disjoint union to its image under the corresponding function $f_\alpha$. This is clearly the only function with the property $f\circ i_\alpha=f_\alpha$ (which, by the way, shows that the disjoint union is also the coproduct in the category of sets). To show that it is continuous, consider the preimage $f^{-1}(U)$ of an open set $U$ of $Y$. The intersections of this preimage with the $X_\alpha$ are precisely the preimages of $U$ under the $f_\alpha$, which are open because the $f_\alpha$ are continuous. Thus the preimage $f^{-1}(U)$ is open, so $f$ is continuous and thus a morphism.
