Building a Map from Local Definitions
A function may be given by different formulas on different parts of its domain. To conclude that the resulting function is continuous, two questions must be answered: do the formulas give the same value wherever their domains overlap, and are those domains open? The open-cover pasting lemma gives a precise answer. Its cover may have any number of pieces, including infinitely many.
Let \((X,\mathcal{T})\) and \((Y,\mathcal{S})\) be topological spaces. A family \(\{U_i:i\in I\}\) is an open cover of \(X\) if every \(U_i\) is open in \(X\) and \(\bigcup_{i\in I}U_i=X\). Each \(U_i\) is given the subspace topology. A map \(f_i:U_i\to Y\) is continuous if the preimage of every open subset of \(Y\) is open in \(U_i\).
The overlap condition is what permits us to define \(f(x)\) using any piece that contains \(x\). Openness is what turns a set open relative to a piece into a set open in the whole domain. Both points appear directly in the proof.
Proof. If \(X=\varnothing\), the empty function \(f:X\to Y\) is the unique map with the required restrictions and is continuous, since the preimage of every set is empty. Now suppose \(X\ne\varnothing\). For \(x\in X\), choose an index \(i\) with \(x\in U_i\), and set \(f(x)=f_i(x)\). If also \(x\in U_j\), the overlap condition gives \(f_i(x)=f_j(x)\), so this definition does not depend on the choice of index. It also follows from the definition that \(f|_{U_i}=f_i\). Any map with these restrictions must take the value \(f_i(x)\) at every \(x\in U_i\), and the \(U_i\) cover \(X\); therefore \(f\) is unique.
To prove continuity, let \(V\in\mathcal{S}\). The definition of \(f\) gives \[ f^{-1}(V)=\bigcup_{i\in I}f_i^{-1}(V). \] For each \(i\), continuity of \(f_i\) says that \(f_i^{-1}(V)\) is open in the subspace \(U_i\). Thus there is an open set \(W_i\in\mathcal{T}\) with \(f_i^{-1}(V)=U_i\cap W_i\). Since \(U_i\) is open in \(X\), \(U_i\cap W_i\) is open in \(X\). The displayed union is therefore open in \(X\), which proves that \(f\) is continuous. \(\square\)
There is also a useful converse. If \(f:X\to Y\) is continuous, then its restrictions to the open sets \(U_i\), with their subspace topologies, are continuous and agree on overlaps. Indeed, for an open \(V\subseteq Y\), the preimage under \(f|_{U_i}\) is \(U_i\cap f^{-1}(V)\), which is open in \(U_i\). Agreement follows because both restrictions take the value \(f(x)\) at each point of an overlap. Consequently, for a map already defined on \(X\), continuity is equivalent to continuity of all its restrictions to the members of an open cover.
Applying the Lemma to Piecewise Maps
Worked Example: A Piecewise Function on the Real Line
Give \(\mathbb{R}\) its standard topology, and define \(U=(-\infty,1)\) and \(V=(0,\infty)\). These sets are open and cover \(\mathbb{R}\): if \(x<1\), then \(x\in U\), and if \(x\geq 1\), then \(x>0\), so \(x\in V\). On these two sets define \[ f_U(x)=0,\qquad f_V(x)=\bigl(\max(0,x-1)\bigr)^2. \]
The map \(f_U\) is constant and continuous. The map \(f_V\) is continuous because \(x\mapsto x-1\), absolute value, and squaring are continuous, and \(\max(0,t)=(t+|t|)/2\). On the overlap \(U\cap V=(0,1)\), we have \(x-1<0\), so \(\max(0,x-1)=0\). Thus \(f_V(x)=0=f_U(x)\) throughout the overlap.
The lemma gives a continuous map on all of \(\mathbb{R}\). Its values are \[ f(x)= \begin{cases} 0,&x<1,\\ (x-1)^2,&x\geq 1. \end{cases} \] At \(x=1\), the second expression gives \((1-1)^2=0\); the point \(1\) belongs to \(V\), not \(U\). For \(x<1\), the \(U\)-formula applies. For \(x>1\), the \(V\)-formula gives \(\bigl(\max(0,x-1)\bigr)^2=(x-1)^2\).
Check: at \(x=2\), the \(V\)-formula gives \(\bigl(\max(0,2-1)\bigr)^2=1\), agreeing with the pasted formula \((2-1)^2=1\). At \(x=1\), it gives \(\bigl(\max(0,1-1)\bigr)^2=0\), agreeing with the value \(0\) on the part to the left.
Worked Example: Pasting a Vector-Valued Map
Let the domain be \(\mathbb{R}\) with its standard topology and the target be \(\mathbb{R}^2\) with its standard topology. Set \(U=(-\infty,2)\), \(V=(1,\infty)\), and define \[ g_U(x)=(x,0),\qquad g_V(x)=\bigl(x,\max(0,x-2)\bigr). \] The open sets \(U,V\) cover \(\mathbb{R}\): every \(x<2\) is in \(U\), and every \(x\geq2\) is in \(V\). Each coordinate of each local map is continuous, so the maps into \(\mathbb{R}^2\) are continuous. In the standard topology on \(\mathbb{R}^2\), this follows because the preimage of a basic open rectangle is the intersection of the open preimages under the two coordinate functions.
The overlap is \(U\cap V=(1,2)\). For \(x\) in this interval, \(x-2<0\), hence \(\max(0,x-2)=0\), and \(g_V(x)=(x,0)=g_U(x)\). The pasted map is therefore continuous, with \[ g(x)= \begin{cases} (x,0),&x<2,\\ (x,x-2),&x\geq2. \end{cases} \] The endpoint \(2\) is covered by \(V\), and the value there is \((2,\max(0,2-2))=(2,0)\).
Check: for \(x=3\), \(g_V(3)=(3,\max(0,3-2))=(3,1)\), which agrees with the second branch \((3,3-2)=(3,1)\). For \(x=1.5\), which lies in the overlap, both local definitions give \((1.5,0)\).
Worked Example: Compatible Maps on a Finite Discrete Space
Let \(X=\{p,q,r\}\) have the discrete topology, in which every subset is open, and let the target be \(\mathbb{R}\) with its standard topology. The sets \(U_1=\{p,q\}\) and \(U_2=\{q,r\}\) form an open cover. Define \[ f_1(p)=0,\quad f_1(q)=2,\qquad f_2(q)=2,\quad f_2(r)=5. \] Every map from a discrete space is continuous: the preimage of any open set is a subset of the domain, and every such subset is open. The overlap is \(\{q\}\), where both maps have value \(2\), so the compatibility condition holds.
The pasted map is \(f(p)=0,\ f(q)=2,\ f(r)=5\), and it is continuous by the lemma. For an explicit check, the open interval \((1,3)\) has preimage \(f^{-1}((1,3))=\{q\}\), which is open in \(X\). In general, the preimage of any open subset of \(\mathbb{R}\) is a subset of \(X\), hence open.
Check: on the overlap point \(q\), the two assigned values are \(f_1(q)=2\) and \(f_2(q)=2\). If the second value had instead been \(3\), there could be no function on \(X\) whose restrictions were both of the altered local maps, because a function cannot assign two values to \(q\).
Why the Hypotheses Matter
Agreement on overlaps is not merely a convenient way to simplify the definition. If the local maps disagree at even one point of an overlap, they cannot be restrictions of a single function on \(X\). The openness condition has a different role: it is used to promote relative openness in each piece to openness in \(X\). The following example shows that continuity on pieces and agreement wherever they overlap do not suffice for an arbitrary cover.
Worked Example: Continuity Can Fail Without an Open Cover
Take \(X=\mathbb{R}\) with its standard topology and let \(A=(-\infty,0]\), \(B=(0,\infty)\). These sets cover \(\mathbb{R}\), but \(A\) is not open in \(\mathbb{R}\). Give both sets their subspace topologies and define \(h_A:A\to\mathbb{R}\) by \(h_A(x)=0\), and \(h_B:B\to\mathbb{R}\) by \(h_B(x)=1\). Both maps are constant, hence continuous. Their overlap is empty, so agreement on the overlap holds vacuously.
The resulting function is \(h(x)=0\) for \(x\leq0\) and \(h(x)=1\) for \(x>0\). But \[ h^{-1}\bigl((-1/2,1/2)\bigr)=(-\infty,0], \] which is not open in \(\mathbb{R}\): every open interval around \(0\) contains positive points and so is not contained in \((-\infty,0]\). Therefore \(h\) is not continuous. The failed hypothesis is that the cover consists of open sets.
Common pitfall. A tempting but invalid argument is: “Each formula is continuous on its piece, so the combined function is continuous.” In the example above, both restrictions are continuous, but the preimage calculation gives \((-\infty,0]\), not an open set. The open-cover lemma does not allow this conclusion because \(A\) is not open. Conversely, a cover may be open and the local maps may each be continuous, but if they disagree on an overlap, they do not define a single pasted function at all. Check the cover and the agreement condition before invoking the lemma.
Check Your Understanding
For each question, identify which part of the open-cover pasting lemma applies.
- Let \(U=(-\infty,0)\) and \(V=(0,\infty)\) in \(\mathbb{R}\). Do they form an open cover of \(\mathbb{R}\)? Explain.
- For \(U=(-\infty,1)\) and \(V=(0,\infty)\), define \(f_U(x)=x\) and \(f_V(x)=x\). Why do these local maps paste to a continuous map?
- Suppose two continuous maps are defined on open sets \(U,V\) covering \(X\), but take different values at some point of \(U\cap V\). Can they be restrictions of one function on \(X\)?
- In the finite discrete example, change \(f_2(q)\) from \(2\) to \(4\). Which hypothesis fails, and why is there no pasted function?
- In the non-open-cover example, calculate the preimage of \((-1/2,1/2)\) and explain why it proves discontinuity.