I'm studying the final topology on a set $X$ induced by a family $\{f_\alpha:X_\alpha\rightarrow X\mid \alpha\in A\}$ and know of its universal property, namely that given a function $g:X\rightarrow Z$, $g$ is continuous iff $g\circ f_\alpha$ is continuous for all $\alpha\in A$. Showing this is not difficult, but it is not constructive.
What bothers me about this is that my usual understanding of universal properties is that they refer to initial or terminal properties and morphisms; typically, they involve the existence of a unique morphism, as the definitions here suggest. As this answer points out, we may use the category of cones along with the forgetful functor from $\mathbf{Top}$ to $\mathbf{Set}$ to show that the above universal property is indeed as the above definitions give. To be honest, I don't know enough category theory to completely understand the process, which may be where my issue lies.
The question I have is then:
How (or if) the universal property of the disjoint union topology can be realized as a special case of the final topology's universal property, since the disjoint union topology is the final topology induced by the canonical injections of $X_\alpha$ into $X=\amalg_{\alpha\in A}{X_\alpha}$, which is constructive in nature, i.e. that it guarantees the existence of a unique morphism such that a certain diagram commutes.