Let $(X_1,\mathcal{T_1})$ , $(X_2,\mathcal{T_2})$ be topological spaces. Define $C=X_1\times\{1\}\cup X_2\times\{2\}$ with the topology $\mathcal{T}$ generated by $\{U\times\{1\}:U\in\mathcal{T_1}\}\cup\{U\times\{2\}:U\in\mathcal{T_2}\}$. Show that $(C,\mathcal{T})$ is a coproduct $X_1\amalg X_2$.
I am just into category theory so I don't know how it is proved. Anyone offer a detailed proof?
To give a quick overview over coproducts first: For two sets $X$ and $Y$, their coproduct is simply their disjoint union $X\sqcup Y=(X\times\{0\})\cup(Y\times\{1\})$ (also written $X+Y$) and you have canonical (injective) embeddings $X\hookrightarrow X+Y,x\mapsto(x,0)$ and $Y\hookrightarrow X+Y,y\mapsto(y,1)$. Their images (basically $X$ and $Y$ respectivly) are a disjoint partition of the coproduct $X+Y$. The coproduct of sets could therefore be described as the smallest set fulfilling the condition, that you can disjointly embed all those sets in it.
In comparison, for the product $X\times Y$ you have canonical (surjective) projection maps $X\times Y\twoheadrightarrow X,(x,y)\mapsto x$ and $X\times Y\twoheadrightarrow Y,(x,y)\mapsto y$. The product of sets could be described as the largest set fulfilling the condition, that you can project it down on all those sets and not map two different elements to the same tuple.
But in category theory, there generally are none of these expressions like "elements", "disjoint" "smallest" or "largest", which contributes to its description as "abstract nonsense". There are only objects and maps, so we need to find a definition only using those.
Defintion (Weak Coproduct) For objects $(X_i)_{i\in I}$ of a category $\mathcal{C}$, an object $X$ with morphisms $(f_i\colon X_i\rightarrow X)_{i\in I}$ is called a weak coproduct.
This already includes a part of the description above, which is that we need to be able to include all objects into their common coproduct. In general, there are many different weak coproducts. Going back to sets, every non-empty set can be a weak coproduct of any family of sets for example. Therefore, this definition is not really helpful yet.
We could encode the expression of the smallest set fulfilling a certain condition, with the fact that it is therefore contained in every other set fulfilling this condition and we can get such an inclusion.
Defintion (Coproduct) A weak coproduct $X$ with morphisms $(f_i\colon X_i\rightarrow X)_{i\in I}$, so that for every other weak coproduct $Y$ with morphisms $(g_i\colon X_i\rightarrow Y)_{i\in I}$, there is a unique morphism (also called universal morphism) $\phi\colon X\rightarrow Y$, so that for each $i\in I$ we have a factorisation $g_i=\psi\circ f_i$, is called a coproduct.
In general, there are many different coproducts, but this time, they are all isomorphic: If we have coproducts $X$ and $Y$, then there are universal morphisms $\phi\colon X\rightarrow Y$ and $\psi\colon Y\rightarrow X$ for comparing $X$ with $Y$ and $Y$ with $X$ according to the above defintion respectivly and therefore $\operatorname{id}_X,\psi\circ\phi\colon X\rightarrow X$ and $\operatorname{id}_Y,\phi\circ\psi\colon Y\rightarrow Y$ are both universal morphisms for comparing $X$ with $X$ and $Y$ with $Y$ respectivly, which results in them being equal because of the uniqueness of the universal morphism, which results in $X$ and $Y$ being isomorphic. Since the coproduct is unique up to isomorphism, we have earned to use the expression $\coprod_{i\in I}X_i$ or $X+Y$.
Now let's use that in the exercise: Denote the canonical (injective) embeddings with $i_1\colon X_1\hookrightarrow C,x_1\mapsto(x_1,1)$ and $i_2\colon X_2\hookrightarrow C,x_2\mapsto(x_2,2)$. Take any topological space $Z$ and continuous maps $f_1\colon X_1\rightarrow Z$ and $f_2\colon X_2\rightarrow Z$. To show, that $C$ is a coproduct, you need to show the existence and uniqueness (don't forget that one) of a continuous map $\phi\colon C\rightarrow Z$, so that $f_1=\phi\circ i_1$ and $f_2=\phi\circ i_2$. This immediately gives you the existence and uniqueness of $\phi$ with $\phi(x_1,1)=f_1(x_1)$ and $\phi(x_2,2)=f_2(x_2)$.
The only thing left to show is, that this $\phi$ is continuous. Here, we need to use the topology of $C$, which results in $i_1$ and $i_2$ being open maps (the topology of $C$ is the coarsest making this possible). Combining this with $f_1=\phi\circ i_1$ and $f_2=\phi\circ i_2$ being continuous shows (For a proof see my answer here.), that $\phi\vert_{\operatorname{img}(i_1)}$ and $\phi\vert_{\operatorname{img}(i_2)}$ are continuous. Since $\operatorname{img}(i_1)$ and $\operatorname{img}(i_2)$ are open as well as closed due to the definition of the topology of $C$, we have that $\phi$ is continuous due to those theorems.
