Show that the disjoint union is the coproduct in the category of topological spaces

Question

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.

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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?

Answer 1

You'd need to check with your lecturer, but I'm a bit dubious about your phrasing "By the characteristic property of the disjoint union space." However you go about stating that characteristic property, it's awfully close to just saying "The characteristic property of the disjoint union is that it's the coproduct." It depends on what other information you have available, but even then for myself I'd prefer to prove continuity of $f$ by checking the pre-image of each open set.

As for uniqueness, that's easy. The only function $f$ such that $f \circ i_{\alpha} = f_{\alpha}$ has to be given on each element $(x, \alpha)$ by $f(x, \alpha)=f(i_{\alpha}(x)) = f_{\alpha}(x)$ so is unique. In hindsight, I'd put that at the start of your proof as the way of identifying $f$ as the function you had to pick.

Answer 2

A typical point of the coproduct is $i_\alpha(x)\in i_\alpha[X_\alpha]$ for some $\alpha$ with $x\in X_\alpha$. Then we must have $f(i_\alpha(x))=f_\alpha(x)$. Therefore there is only one possible value for each $f(y)$ for $y\in\coprod_\alpha X_\alpha$.

Answer 3

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.

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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.