Maps Out of a Disjoint Union
A disjoint union lets us assemble spaces while keeping their points separate. Its universal property says that to define a map from the assembled space, it is enough to define a map on each component. Moreover, continuity of the assembled map can be checked component by component. The useful feature is that there is no compatibility condition between different components: they have no points in common.
Let \(J\) be an index set and let \((X_\alpha,\mathcal{T}_\alpha)\) be a topological space for each \(\alpha\in J\). We use a tagged version of the disjoint union, so that even if two component spaces contain the same underlying point, their copies remain distinct.
In particular, each point of the disjoint union belongs to exactly one tagged component. The inclusion \(\iota_\alpha\) is continuous: for an open set \(U\) in the union, \(\iota_\alpha^{-1}(U)=U_\alpha\), which is open in \(X_\alpha\). Each component image is also open in the union, since its slice in its own component is all of that component and its other slices are empty. These facts follow directly from the stated topology.
The Continuity Test
Proof. Every point of the union has the form \((\alpha,x)\), so the equation \(f\circ\iota_\alpha=f_\alpha\) forces \(f(\alpha,x)=f_\alpha(x)\). This defines a map, and it is unique because it specifies the value at every point. Suppose first that \(f\) is continuous. For each open \(V\subseteq Y\), the set \(f_\alpha^{-1}(V)\) equals \(\iota_\alpha^{-1}(f^{-1}(V))\). Since \(f^{-1}(V)\) is open and \(\iota_\alpha\) is continuous, this set is open in \(X_\alpha\). Thus each \(f_\alpha\) is continuous. Conversely, suppose each \(f_\alpha\) is continuous, and take any open \(V\subseteq Y\). The \(\alpha\)-slice of \(f^{-1}(V)\) is exactly \(f_\alpha^{-1}(V)\), which is open in \(X_\alpha\). By the definition of the disjoint-union topology, \(f^{-1}(V)\) is open. Hence \(f\) is continuous. If \(J\) is empty, the union is empty and there is exactly one map from it to \(Y\); it is continuous, and the condition on the family is vacuous. \(\square\)
The theorem is both a construction rule and a test. A family of maps specifies one and only one map on the union, whether or not the component maps are continuous. Continuity is the additional property that holds precisely when each component map is continuous.
Proof. Send a map \(f\) to the family \((f\circ\iota_\alpha)_{\alpha\in J}\). In the other direction, send a family \((f_\alpha)\) to the unique map defined by \(f(\alpha,x)=f_\alpha(x)\). The construction in the theorem shows that starting with a family and restricting the assembled map returns that same family. Starting with \(f\) and assembling its restrictions returns \(f\), since at every tagged point \((\alpha,x)\) the assembled map has value \(f(\iota_\alpha(x))=f(\alpha,x)\). The theorem's continuity test gives the final assertion. \(\square\)
Constructing Maps One Component at a Time
Worked Example: Two Interval Components Mapping to the Real Line
Let \(X_1\) and \(X_2\) each be a copy of \(I=[0,1]\) with the usual subspace topology from \(\mathbb{R}\). Define \(f_1:X_1\to\mathbb{R}\) by \(f_1(t)=t\), and \(f_2:X_2\to\mathbb{R}\) by \(f_2(t)=2+t\), where \(\mathbb{R}\) has its standard topology. Construct the corresponding map \(f:X_1\bigsqcup X_2\to\mathbb{R}\).
The universal property gives \(f(1,t)=t\) and \(f(2,t)=2+t\). The first restriction is continuous because \(|f_1(t)-f_1(s)|=|t-s|\). The second is continuous because \(|f_2(t)-f_2(s)|=|(2+t)-(2+s)|=|t-s|\). Thus the assembled map is continuous.
At the endpoints, \(f(1,0)=0\), \(f(1,1)=1\), \(f(2,0)=2\), and \(f(2,1)=3\). The first component maps onto \([0,1]\); the second maps onto \([2,3]\). As a further check, take the open target set \(V=(1/4,1/2)\). On the first component its preimage is \((1/4,1/2)\). On the second component the condition \(1/4<2+t<1/2\) would require \(-7/4<t<-3/2\), which has no solutions in \([0,1]\). Therefore \(f^{-1}(V)\) consists of the first component's tagged interval \(\{(1,t):1/4<t<1/2\}\) and no points of the second component. This is open in the disjoint union, as the slice in each component is open.
Check: the formula on the first copy gives \(f(1,1/2)=1/2\), while the formula on the second gives \(f(2,1/2)=2+1/2=5/2\). These are distinct values at distinct tagged points, and both formulas agree with the stated ranges.
Worked Example: A Map from Countably Many Singletons
For each \(n\in\mathbb{N}\), let \(X_n=\{x_n\}\) with its unique topology, and let the target be \(\mathbb{R}\) with its standard topology. Define \(f_n(x_n)=1/n\). The unique map \(f:\bigsqcup_{n\in\mathbb{N}}X_n\to\mathbb{R}\) therefore satisfies \(f(n,x_n)=1/n\).
Each \(f_n\) is continuous: the preimage of any open subset of \(\mathbb{R}\) is either the empty set or \(X_n\), both open in the singleton space. Hence \(f\) is continuous. In fact, every subset of this disjoint union is open, because its slice in each singleton is either empty or the whole singleton. So any assignment of real values to these tagged points would give a continuous map.
For a direct check, let \(V=(1/4,1/2)\). A point \((n,x_n)\) belongs to \(f^{-1}(V)\) exactly when \(1/4<1/n<1/2\). Multiplying the positive inequalities by \(4n\) and \(2n\), respectively, gives \(n<4\) and \(n>2\). The only positive integer satisfying both is \(n=3\), and \(f(3,x_3)=1/3\) lies strictly between \(1/4\) and \(1/2\). Thus \(f^{-1}(V)=\{(3,x_3)\}\), which is open in this disjoint union.
Check: \(f(1,x_1)=1\), \(f(2,x_2)=1/2\), and \(f(3,x_3)=1/3\). The first two values do not lie in \(V\), while the third does, exactly as the computed preimage states.
Worked Example: Maps into the Sierpinski Space
Let \(X_1=\{a,b\}\) have topology \(\{\varnothing,\{a\},X_1\}\), and let \(X_2=I=[0,1]\) have its usual topology. Let \(Y=\{a,b\}\) have the same topology as \(X_1\). Define \(f_1:X_1\to Y\) to be the identity and \(f_2:X_2\to Y\) to be the constant map with value \(a\).
The identity \(f_1\) is continuous because the preimage of each open set is that same open set. The constant map \(f_2\) is continuous because the preimage of an open set containing \(a\) is all of \(X_2\), and the preimage of the only open set not containing \(a\), namely \(\varnothing\), is empty. Both \(X_2\) and \(\varnothing\) are open in \(X_2\).
The assembled map is \(f(1,a)=a\), \(f(1,b)=b\), and \(f(2,t)=a\) for every \(t\in[0,1]\). To check its continuity directly, the open sets of \(Y\) are \(\varnothing,\{a\},Y\). Their preimages are, respectively, \(\varnothing\), \(\{(1,a)\}\cup\{(2,t):t\in[0,1]\}\), and the whole disjoint union. The middle preimage has slice \(\{a\}\) in \(X_1\), which is open, and slice \(X_2\) in \(X_2\), which is open. All three preimages are therefore open.
Check: the middle preimage excludes \((1,b)\), since \(f(1,b)=b\), and includes every \((2,t)\), since \(f(2,t)=a\). This agrees with the full list of values of \(f\).
A Common Pitfall: Requiring Components to Match
One mistake is to think that formulas on different components must agree wherever their input coordinates look the same. For the two interval copies in the first example, the coordinate \(1/2\) appears in both, but the domain points are \((1,1/2)\) and \((2,1/2)\). Their values are \(1/2\) and \(5/2\), respectively. They are different domain points, so assigning them different values is not a failure to define a function. A matching condition would be relevant only if the domain identified those points; the disjoint union does not.
The opposite error is to suppose every map from a disjoint union is continuous. Let the domain be the disjoint union of two copies of \(\mathbb{R}\), each with the standard topology, and let the target also be standard \(\mathbb{R}\). On the first copy define the constant-zero map. On the second define \(h(x)=-1\) for \(x<0\), \(h(0)=0\), and \(h(x)=1\) for \(x>0\). The preimage \(h^{-1}((-1/2,1/2))\) is \(\{0\}\), which is not open in \(\mathbb{R}\). Thus \(h\) is not continuous, and the assembled map is not continuous: its preimage of \((-1/2,1/2)\) has a singleton slice in the second component, so it is not open in the disjoint union. The componentwise test requires continuity on every component, not merely a formula on each one.
The universal property is useful whenever a domain is built as a collection of separate pieces. It turns a global construction into independent choices of component maps, and its continuity test proves that no additional cross-component condition is needed.
Check Your Understanding
Use the definition of the disjoint-union topology and the continuity test to answer these questions.
- For maps \(f_\alpha:X_\alpha\to Y\), write the unique formula for the induced map from \(\bigsqcup_{\alpha\in J}X_\alpha\) to \(Y\).
- Explain why continuity of the induced map implies continuity of every component map.
- Suppose \(J\) is empty. What is the disjoint union, and how many maps does it have to a fixed space \(Y\)?
- Let \(X_1=X_2=\mathbb{R}\) with their standard topologies, and define \(f(1,x)=x\) and \(f(2,x)=x+1\). Is \(f\) continuous? Justify your answer from the component maps.
- In the countable-singleton example, determine \(f^{-1}((1/3,1))\), taking care about the strict endpoints.