Continuity After Gluing
Many spaces are built by taking several pieces and identifying selected points. A map out of the resulting space can be described on each piece, but continuity is not checked by treating the pieces as separate copies: the descriptions must agree at every identification. Quotient topology gives an exact test. The disjoint-union topology then reduces the test to continuity on each piece.
For a space \((X,\mathcal{T})\), a surjection \(q:X\to Y\) makes \(Y\) a quotient space of \(X\) when its topology is defined by the rule that \(V\subseteq Y\) is open exactly when \(q^{-1}(V)\) is open in \(X\). Such a map \(q\) is called a quotient map. The topology on \(Y\) is part of this definition; it is not generally the topology \(Y\) would inherit as a subset of another space.
Proof. Let \(W\) be any open subset of \(Z\). The definition of continuity says that \(g\) is continuous exactly when \(g^{-1}(W)\) is open in \(Y\). Since \(q\) is a quotient map, this is equivalent to \(q^{-1}(g^{-1}(W))\) being open in \(X\). The preimage identity \(q^{-1}(g^{-1}(W))=(g\circ q)^{-1}(W)\) says that this last condition holds for every open \(W\subseteq Z\) exactly when \(g\circ q\) is continuous. \(\square\)
The quotient-map hypothesis matters: the test converts openness in \(Y\) into openness after taking a preimage under \(q\). Surjectivity by itself does not provide that conversion.
Maps Defined on Disjoint Pieces
Suppose \(\{X_i\}_{i\in J}\) is a family of topological spaces, and let \(D=\bigsqcup_{i\in J}X_i\) be their disjoint union. Its topology is defined so that \(U\subseteq D\) is open if and only if \(U\cap X_i\) is open in \(X_i\) for every \(i\). Each inclusion \(\iota_i:X_i\to D\) is continuous: the preimage \(\iota_i^{-1}(U)=U\cap X_i\) is open whenever \(U\) is open in \(D\).
Proof. If \(h\) is continuous, then for each \(i\), the restriction \(h\circ\iota_i\) is continuous because both \(h\) and \(\iota_i\) are continuous. Conversely, suppose every restriction is continuous. For an open set \(W\subseteq Z\), we have \((h^{-1}(W))\cap X_i=(h\circ\iota_i)^{-1}(W)\), which is open in \(X_i\) for every \(i\). By the definition of the disjoint-union topology, \(h^{-1}(W)\) is open in \(D\). This holds for every open \(W\), so \(h\) is continuous. \(\square\)
Now let \(q:D\to Y\) be a quotient map that performs a gluing. Points in \(D\) may be identified by \(q\). A family of maps \(h_i:X_i\to Z\) defines a map on the glued space only if it assigns the same value to any two points identified by \(q\). When that compatibility holds, there is a unique map \(g:Y\to Z\) with \(g\circ q|_{X_i}=h_i\). The quotient continuity test and the disjoint-union test together give a practical criterion.
Proof. Compatibility makes \(g\) well-defined: for \(y\in Y\), choose \(x\in D\) with \(q(x)=y\) and set \(g(y)=h(x)\), where \(h\) is the map on \(D\) specified by the \(h_i\). If another representative \(x'\) has \(q(x')=y\), compatibility gives \(h(x')=h(x)\), so the value does not depend on the choice. The disjoint-union test says that \(h:D\to Z\) is continuous exactly when all \(h_i\) are continuous. Since \(h=g\circ q\), the quotient continuity test says that \(h\) is continuous exactly when \(g\) is continuous. These two equivalences prove the criterion. For necessity, if an induced \(g\) is continuous, then \(h_i=g\circ q\circ\iota_i\) is continuous for each \(i\), and equal images under \(q\) must have equal values under \(g\). \(\square\)
Worked Uses of the Criterion
Worked Example: Joining Two Intervals
Take two copies \(I_1,I_2\) of \(I=[0,1]\), each with the subspace topology from \(\mathbb{R}\). Give \(D=I_1\bigsqcup I_2\) the disjoint-union topology, and form the quotient \(q:D\to Y\) by identifying the point \(1\) in \(I_1\) with the point \(0\) in \(I_2\), with no other identifications. Define \(h_1:I_1\to\mathbb{R}\) by \(h_1(t)=t\), and \(h_2:I_2\to\mathbb{R}\) by \(h_2(t)=1+t\). Both are continuous, since each is the restriction of a continuous affine function on \(\mathbb{R}\).
At the identified points, the two values are \(h_1(1)=1\) and \(h_2(0)=1+0=1\). Thus the maps are compatible and induce \(g:Y\to\mathbb{R}\). At a point represented by \(t\in I_1\), \(g\) has value \(t\); at a point represented by \(t\in I_2\), it has value \(1+t\). The criterion proves \(g\) is continuous.
Check: the glued point has value \(1\) whichever representative is used. At the midpoint of \(I_1\), the value is \(h_1(1/2)=1/2\); at the midpoint of \(I_2\), it is \(h_2(1/2)=1+1/2=3/2\). These calculations agree with the formulas on the respective pieces and confirm that the endpoint identification, not equality of the two formulas everywhere, is what compatibility requires.
If instead \(h_2(t)=2+t\), then \(h_1(1)=1\) but \(h_2(0)=2+0=2\). No map on \(Y\) can induce those two formulas: the single quotient point would have to be sent to both \(1\) and \(2\).
Worked Example: A Map from a Mapping Cylinder
Let \(f:I\to I\) be \(f(x)=x^2\), and give both intervals their standard topology. The mapping cylinder \(M_f\) is the quotient of \(D=(I\times I)\bigsqcup I\) that identifies \((x,1)\) in the first piece with \(f(x)\) in the second piece for each \(x\in I\). The product \(I\times I\) has the product topology, and \(D\) has the disjoint-union topology.
Define \(a:I\times I\to I\) by \(a(x,t)=(1-t)x+tx^2\), and define \(b:I\to I\) by \(b(y)=y\). The formula for \(a\) is polynomial in \(x,t\), so \(a\) is continuous; \(b\) is the identity and is continuous. For each \(x\in I\), the values on the points to be identified are \(a(x,1)=(1-1)x+1\cdot x^2=x^2\) and \(b(f(x))=b(x^2)=x^2\). Thus the formulas are compatible and induce \(G:M_f\to I\). The gluing criterion proves \(G\) is continuous.
For a numerical check, at \(x=1/2\) and \(t=1/3\), \(a(1/2,1/3)=(1-1/3)(1/2)+(1/3)(1/2)^2=(2/3)(1/2)+(1/3)(1/4)=1/3+1/12=5/12\). At the identified top point \(t=1\), the same formula gives \(a(1/2,1)=(0)(1/2)+(1)(1/4)=1/4\), and the corresponding point \(f(1/2)=1/4\) in the second piece has \(b(1/4)=1/4\). The endpoint check verifies compatibility in this instance.
Why Quotient Topology Is Not Subspace Topology
A quotient topology is determined by which subsets have open preimages under the quotient map. It is generally not the topology obtained by viewing a set of representatives as a subset of the original space. The following example shows the distinction explicitly.
Worked Example: Identifying the Endpoints of an Interval
Let \(I=[0,1]\) have its standard topology, and let \(q:I\to Y\) identify \(0\) with \(1\), leaving all other points distinct. Give \(Y\) the quotient topology. Each equivalence class has a unique representative in \([0,1)\); hence there is a bijection \(r:Y\to[0,1)\) sending each class to that representative. Give \([0,1)\) the subspace topology from \(\mathbb{R}\).
The set \(U=[0,1/4)\) is open in \([0,1)\), since \(U=(-1/8,1/4)\cap[0,1)\). Its preimage under \(r\circ q\) is \((r\circ q)^{-1}(U)=[0,1/4)\cup\{1\}\). This subset is not open in \(I\): it contains \(1\), but every open neighborhood of \(1\) in \(I\) contains points less than \(1\), and points sufficiently close to \(1\) are not in \([0,1/4)\cup\{1\}\). Therefore \(r\circ q\) is not continuous.
If \(r\) were continuous, then \(r\circ q\) would be continuous as a composition of continuous maps, contradicting the preimage computation. Thus the bijection from the quotient to its chosen representatives is not a homeomorphism. In particular, the quotient topology cannot simply be replaced by the subspace topology on \([0,1)\).
A Common Error: Surjective Is Not the Same as Quotient
Common pitfall. One may try to use the quotient continuity test whenever a map is surjective. The required hypothesis is that the map be a quotient map, not merely a surjection. For a concrete failure, let \(D=\{0,1\}\) have the discrete topology and let \(Y=\{0,1\}\) have the indiscrete topology \(\{\varnothing,Y\}\). The identity function \(q:D\to Y\) is continuous and surjective. Let \(Z=\{0,1\}\) have the discrete topology and let \(g:Y\to Z\) be the identity function.
The composite \(g\circ q:D\to Z\) is continuous: every subset of \(D\) is open, so the preimage of every open subset of \(Z\) is open. But \(g\) is not continuous, because \(g^{-1}(\{0\})=\{0\}\) is not open in the indiscrete space \(Y\). Correspondingly, \(q\) is not a quotient map: the subset \(\{0\}\subseteq Y\) has open preimage in \(D\) but is not open in \(Y\). The quotient topology induced by \(q\) would be discrete, not indiscrete.
The quotient test is useful precisely because it replaces an open-set check in a potentially complicated glued space with checks on the pieces. Its two essential steps are: first verify that the maps agree at every identification, and then verify continuity on each piece. Compatibility gives a well-defined function; the quotient topology gives the continuity conclusion.
Check Your Understanding
For each question, identify the relevant topology and check both continuity and compatibility where appropriate.
- In the endpoint quotient of \(I\), define a real-valued function on \(I\) by \(h(t)=t\). Does it descend to a function on the quotient? Explain using the values at \(0\) and \(1\).
- For two copies of \(I\) with \(1\) in the first identified with \(0\) in the second, test whether \(h_1(t)=t^2\) and \(h_2(t)=t+1\) induce a continuous map to \(\mathbb{R}\).
- State exactly which assumption on \(q:X\to Y\) is needed in the quotient continuity test, and why surjectivity alone does not suffice.
- For \(f:I\to I\) defined by \(f(x)=x^2\), check compatibility of \(a(x,t)=(1-t)x+tx^2\) and \(b(y)=y\) at every identified pair \((x,1)\sim f(x)\).
- In the endpoint quotient example, why is \([0,1/4)\cup\{1\}\) not open in \(I\), even though its image among representatives corresponds to an open subset of \([0,1)\)?