Given any family $\{X_{\alpha}\}_{\alpha \in A}$ of topological spaces, show that the disjoint union space $\bigsqcup_{\alpha \in A} X_\alpha$ is their coproduct in the category Top
My attemtped proof: Fix $\alpha \in A$. The disjoint union space comes together with a continuous inclusion map $i_{\alpha} : X_{\alpha} \to \bigsqcup_{\alpha \in A}X_{\alpha}$. Let $W$ be any topological space and let $f_{\alpha} : X_{\alpha} \to W$ be any continuous map.
We need to show that there exists a unique continuous map $f: \bigsqcup_{\alpha \in A}X_{\alpha} \to W$ such that $f \circ i_{\alpha} = f_{\alpha}$. To that end define $f : \bigsqcup_{\alpha \in A}X_{\alpha} \to W$ by $$f(x, \alpha) = f_{\alpha}(x).$$By the characteristic property of the disjoint union space, $f$ is continuous if and only if its restriction to each $(X_{\alpha})^* = i_{\alpha}[X_{\alpha}]$ is continuous. So it suffices to show that such a restriction is continuous to show continuity of $f$.
With that in mind consider the restriction $f|_{(X_{\alpha})^*} : (X_{\alpha})^* \to W$ and define $\pi_{\alpha} : i_{\alpha}[X_{\alpha}] \to X_{\alpha}$ by $\pi(x, \alpha) = x$. Note that $\pi_{\alpha}$ is continuous and we have $$f|_{(X_{\alpha})^*} = f_{\alpha} \circ \pi_{\alpha}$$ and since $f_\alpha $ and $\pi_\alpha$ are continuous we have that $f|_{(X_{\alpha})^*}$ is continuous. Since $\alpha$ was fixed arbitrarily this holds for all $\alpha \in A$. Hence by the characteristic property we conclude that $f$ is continuous.
Now I have to show that the map $f$ is unique. However I'm not sure how to proceed with that. How can I prove that $f$ is unique? Also is what I've written above correct?
Best Answer
This is a very good exercise to do it by hand as you attempted to and others have already provided some help. I just want to complete these by proposing a very "abstract nonsense" proof.
The forgetful functor $U : \mathbf{Top} \to \mathbf{Set}$ admits a right adjoint, namely the functor that sends a set $X$ to the space $X$ with trivial topology (some call it indiscrete topology also i think). So the forgetful functor $U$ commutes with colimits, meaning in particular that the underlying set of the disjoint union of topological spaces $(X_\alpha)_{\alpha\in A}$ is the disjoint union of the underlying sets $(U(X_\alpha))_{\alpha\in A}$. Now you just have to find the topology you need to put on $\coprod_\alpha U(X_\alpha)$ in order to make it the disjoint union in $\mathbf{Top}$. But the open sets of any topological space $X$ are identified with the continuous map $X\to \mathbb S$, where $\mathbb S$ is the Sierpinski space : it has two elements $0$ and $1$ with $\{0\}$ open and $\{1\}$ not open. In particular, whatever $\coprod_\alpha X_\alpha$ is, if it exists, its set of open sets is in bijection with $$\mathbf{Top}(\coprod_\alpha X_\alpha,\mathbb S)\simeq \prod_\alpha\mathbf{Top}(X_\alpha,\mathbb S)$$ The right hand side is just a use of the universal property of the coproduct. Hence it says that an open set of $\coprod_\alpha X_\alpha$ is exactly given by an open set in each of the $X_\alpha$'s. Now you have to be a little careful: the bijection just above is given by precomposition with the canonical maps $\iota_\beta:X_\beta \to \coprod_\alpha X_\alpha$. So the bijection is actually saying that $\coprod_\alpha X_\alpha$ has the final topology given by the maps $U(\iota_\beta)$. Taking into account the remark we made about $U(\coprod_\alpha X_\alpha)$, it is actually the definition of the disjoint union topology.