The Gluing Setup
An adjunction space is formed by taking two spaces and identifying points of one with their prescribed images in the other. The quotient topology is important: it makes continuity of a map out of the glued space testable on the original pieces. The result is a precise universal property, not just a useful way to describe particular maps.
Let \(X\) and \(Y\) be topological spaces, let \(A\subseteq X\) have the subspace topology, and let \(f:A\to Y\) be continuous. Write \(i:A\to X\) for the inclusion map, \(i(a)=a\). Give the disjoint union \(X\bigsqcup Y\) its disjoint-union topology: a subset is open exactly when its intersection with each piece is open in that piece.
The equivalence relation may identify more than one pair of points: for example, different elements of \(A\) can have the same image under \(f\). Its definition ensures that \(j_X(i(a))=j_Y(f(a))\) for every \(a\in A\). No injectivity of \(f\), or of either map \(j_X,j_Y\), is assumed. If \(A=\varnothing\), there are no identifications and \(P=X\bigsqcup Y\).
The Pushout Property
The equality \(u\circ i=v\circ f\) means that the two maps assign the same value to every pair of points identified in forming \(P\). This is the compatibility condition. The theorem says that the compatible pair determines exactly one map on the glued space.
Proof. First suppose \(g:P\to Z\) is continuous. The maps \(j_X\) and \(j_Y\) are continuous: each is the composite of the inclusion of one piece into the disjoint union and the quotient map \(q\). Therefore \(u=g\circ j_X\) and \(v=g\circ j_Y\) are continuous. For each \(a\in A\), the defining identification gives \(u(i(a))=g(j_X(i(a)))=g(j_Y(f(a)))=v(f(a))\). Thus \(u\circ i=v\circ f\).
Conversely, suppose \(u\) and \(v\) are continuous and satisfy that equality. Define \(h_0:X\bigsqcup Y\to Z\) by \(h_0(x)=u(x)\) on the \(X\) piece and \(h_0(y)=v(y)\) on the \(Y\) piece. This map is continuous: for any open \(W\subseteq Z\), the intersections of \(h_0^{-1}(W)\) with the two pieces are \(u^{-1}(W)\) and \(v^{-1}(W)\), both open in their respective spaces. The compatibility equality says that \(h_0\) takes equal values on every generating pair \(i(a),f(a)\). It therefore takes equal values along every finite chain of generating identifications, and hence is constant on each equivalence class. So there is a well-defined map \(g:P\to Z\) with \(g\circ q=h_0\).
To check continuity, let \(W\subseteq Z\) be open. By the definition of the quotient topology, \(g^{-1}(W)\) is open in \(P\) exactly when \(q^{-1}(g^{-1}(W))\) is open in \(X\bigsqcup Y\). But \(q^{-1}(g^{-1}(W))=h_0^{-1}(W)\), which is open because \(h_0\) is continuous. Thus \(g\) is continuous. Its restrictions satisfy \(g\circ j_X=u\) and \(g\circ j_Y=v\). Finally, any map satisfying these two equations must agree with \(g\) on the images of both pieces, whose union is \(P\); hence it must equal \(g\). \(\square\)
This proof uses two defining features of the construction. The disjoint-union topology reduces continuity to continuity on each piece, and the quotient topology says that continuity of a map out of \(P\) can be checked after composing with \(q\). In particular, the theorem does not require \(j_X\) or \(j_Y\) to be embeddings.
Worked Applications
Worked Example: Attaching a Point to an Interval
Let \(X=I=[0,1]\), \(A=\{0\}\) with its subspace topology, \(Y=\{p\}\) with its unique topology, and \(f(0)=p\). The quotient identifies \(0\in I\) with \(p\). Take \(Z=\mathbb{R}\) with its standard topology, and define \(u:I\to\mathbb{R}\) by \(u(x)=x^2\) and \(v:Y\to\mathbb{R}\) by \(v(p)=0\). Both maps are continuous, and the compatibility check is \(u(i(0))=u(0)=0=v(p)=v(f(0))\). The theorem gives a unique continuous \(g:P\to\mathbb{R}\).
On the quotient classes, its values are \(g(j_X(x))=x^2\) for \(x\in I\), and \(g(j_Y(p))=0\). These formulas agree on the only class containing points from both pieces, since \(j_X(0)=j_Y(p)\) and both assigned values are \(0\). For a numerical check, at \(x=1/2\) the value is \((1/2)^2=1/4\), while the identified point has value \(0^2=0\). Thus the formula respects the identification and has the required restrictions.
In fact, this particular quotient is homeomorphic to \(I\). The map \(r:I\bigsqcup\{p\}\to I\) that is the identity on \(I\) and sends \(p\) to \(0\) is continuous by the disjoint-union topology. It is constant on equivalence classes, so it induces a continuous map \(\bar r:P\to I\). The map \(j_X:I\to P\) is continuous, and \(\bar r\circ j_X=\operatorname{id}_I\). Also every point of \(P\) is \(j_X(x)\) for some \(x\in I\), because \(j_Y(p)=j_X(0)\). Hence \(j_X\) and \(\bar r\) are inverse continuous maps.
Worked Example: Two Endpoint Identifications
Take \(X=I\), \(A=\{0,1\}\) with its subspace topology, \(Y=I\), and define \(f:A\to Y\) by \(f(0)=1\) and \(f(1)=0\). The adjunction space identifies the first interval’s \(0\) with the second interval’s \(1\), and the first interval’s \(1\) with the second interval’s \(0\).
Let \(Z=I\), and set \(u(x)=x\) on \(X\) and \(v(y)=1-y\) on \(Y\). These are continuous. Check both required equations: at \(a=0\), \(u(i(0))=u(0)=0\) and \(v(f(0))=v(1)=1-1=0\); at \(a=1\), \(u(i(1))=u(1)=1\) and \(v(f(1))=v(0)=1-0=1\). Thus the theorem gives a unique continuous map \(g:P\to I\).
For example, the class containing the first interval’s \(0\) and the second interval’s \(1\) receives value \(0\) from either formula. The class containing the first interval’s \(1\) and the second interval’s \(0\) receives value \(1\) from either formula. At the second interval’s point \(y=1/2\), the value is \(v(1/2)=1-1/2=1/2\). These checks confirm that the piecewise formulas define one map on the quotient, not conflicting values on identified points.
Worked Example: Collapsing Both Endpoints to One Point
Let \(X=[-1,1]\) with its standard subspace topology from \(\mathbb{R}\), let \(A=\{-1,1\}\), let \(Y=\{p\}\), and let \(f\) send both elements of \(A\) to \(p\). The quotient identifies \(-1\), \(1\), and \(p\) in one equivalence class.
Define \(u:X\to\mathbb{R}\) by \(u(x)=1-x^2\), and define \(v(p)=0\). The polynomial formula makes \(u\) continuous, and \(v\) is continuous. For each attaching point, the compatibility calculation is \(u(-1)=1-(-1)^2=1-1=0=v(p)\) and \(u(1)=1-1^2=1-1=0=v(p)\). The theorem therefore gives a unique continuous \(g:P\to\mathbb{R}\).
The resulting map has \(g(j_X(x))=1-x^2\). At \(x=1/2\), this gives \(1-(1/2)^2=1-1/4=3/4\); at \(x=-1/2\), it also gives \(1-(-1/2)^2=1-1/4=3/4\). At either endpoint the value is \(0\), agreeing with the value at \(j_Y(p)\). The endpoint checks are exactly what makes the formula well-defined on the collapsed class.
What the Hypotheses Prevent
Common pitfall: checking agreement but not continuity. Keep the setup of the first example, where \(I\)’s \(0\) is identified with \(p\). Define \(u(0)=0\), \(u(x)=1\) for \(x>0\), and \(v(p)=0\). The compatibility check at the identified point succeeds: \(u(0)=v(p)=0\). But \(u\) is not continuous at \(0\): the inverse image \(u^{-1}((-1/2,1/2))\) is \(\{0\}\), which is not open in \(I\). Consequently, these formulas induce a set map on the quotient, but not a continuous map. The correction is to check both requirements in the theorem: continuity on each piece and agreement on the attaching data.
Common pitfall: checking only one identification. In the endpoint example with \(f(0)=1\) and \(f(1)=0\), the formulas \(u(x)=x\) and \(v(y)=0\) agree at the first identification, since \(u(0)=0=v(1)\). They fail at the second: \(u(1)=1\), whereas \(v(0)=0\). The two points of that pair represent the same quotient point, so assigning them different values cannot define a map on \(P\). Compatibility must be checked for every \(a\in A\).
The universal property is useful because it turns a problem about a quotient into a problem about its pieces. To construct a continuous map out of an adjunction space, define continuous maps on \(X\) and \(Y\), verify their values agree under the attaching map, and then invoke the theorem. Conversely, any continuous map out of the quotient necessarily restricts to such a compatible pair. The theorem also identifies the precise sense in which \(P\) is the pushout of the inclusion \(i:A\to X\) and the attaching map \(f:A\to Y\): every compatible pair of maps from \(X\) and \(Y\) to a target \(Z\) factors uniquely through \(P\).
Check Your Understanding
For each question, identify the attaching data and test the compatibility condition at every element of \(A\).
- If \(A=\varnothing\), what is the adjunction space, and what does the pushout property say about maps from it?
- Let \(A=\{0,1\}\subseteq I\), let \(f(0)=f(1)=p\) for \(Y=\{p\}\), and take \(Z=\mathbb{R}\). What conditions must a continuous \(u:I\to\mathbb{R}\) and a map \(v:Y\to\mathbb{R}\) satisfy to induce a map from the quotient?
- In the two-endpoint example, replace \(v(y)=1-y\) by \(v(y)=y\), keeping \(u(x)=x\). At which attaching points does compatibility hold or fail?
- Why does a continuous map \(g:P\to Z\) give continuous restrictions \(g\circ j_X\) and \(g\circ j_Y\)?
- In the collapsed-endpoints example, why would \(u(x)=x\) and \(v(p)=0\) fail to define any map on the quotient, even before continuity is considered?