Attaching a Cylinder to a Target
A continuous map \(f:X\to Y\) can be incorporated into a space by taking a cylinder on \(X\) and attaching one end to \(Y\). The resulting space retains a copy of the target and a copy of the source at the other end of the cylinder. The quotient topology is essential: it specifies which subsets of the glued space are open, including near the attached end.
Throughout, \(X\) and \(Y\) are topological spaces, with their stated topologies, and \(I=[0,1]\) has its usual topology. The product \(X\times I\) has the product topology, and \(Y\bigsqcup(X\times I)\) has the disjoint union topology: a subset is open exactly when its intersections with both pieces are open. Assume \(f:X\to Y\) is continuous.
Only points at the bottom of the cylinder are identified with points of \(Y\). More explicitly, the equivalence class of \(y\in Y\) consists of \(y\) together with all \((x,0)\) for which \(f(x)=y\). Every \((x,t)\) with \(t>0\) is in a singleton class. In particular, no points at the top \(X\times\{1\}\) are identified with one another or with points of \(Y\).
Reading Openness from the Quotient
Proof. A saturated subset must contain \((x,0)\) exactly when it contains its equivalent point \(f(x)\). Thus its bottom-cylinder points are precisely \((x,0)\) with \(x\in f^{-1}(U)\), which is the displayed condition. Conversely, if this condition holds, then whenever \(G\) contains one member of an equivalence class, it contains every member: this is exactly the required matching of \(y\) and all bottom points mapped to \(y\). The openness assertion is the definition of the quotient topology. In particular, if \(G\) is both open and saturated, then \(q(G)\) is open, because \(q^{-1}(q(G))=G\). \(\square\)
This criterion separates two tasks. First specify an open set in each piece of the disjoint union; then check that membership at the bottom of the cylinder agrees with membership in the target. A set in the disjoint union that fails this matching condition cannot be the preimage of any subset of \(M_f\).
Worked Example: The Identity Map on a One-Point Space
Let \(X=Y=\{p\}\), each with its unique topology, and let \(f(p)=p\). The disjoint union has one target point and a copy of \(I\). The quotient identifies the target point with the cylinder point \((p,0)\); all \((p,t)\) for \(t>0\) remain distinct. Thus the map \(h:I\to M_f\) given by \(h(t)=q(p,t)\) is a bijection, where \(h(0)=q(p)=j(p)\).
To check the topology, let \(O\subseteq M_f\) and put \(V=\{t\in I:h(t)\in O\}\). The cylinder part of \(q^{-1}(O)\) is \(\{p\}\times V\), and its target part contains \(p\) exactly when \(0\in V\). The target singleton is open in its component. Therefore \(q^{-1}(O)\) is open exactly when \(V\) is open in \(I\). By the quotient definition, \(O\) is open exactly when \(h^{-1}(O)=V\) is open. Hence \(h\) is a homeomorphism from \(I\) to \(M_f\).
Check: under this correspondence the attached target point is \(h(0)\), while the canonical source point is \(i(p)=h(1)\). They are different, since \(0\ne1\) and no relation identifies \((p,1)\) with the target.
Worked Example: Two Source Points Attached to One Target Point
Let \(X=\{a,b\}\) have the discrete topology, let \(Y=\{y\}\) have its unique topology, and let \(f(a)=f(b)=y\). The quotient set has the point \(y\), together with distinct points represented by \((a,t)\) and \((b,t)\) for \(0<t\leq1\). Both bottom points represent \(y\); points on the two branches at positive heights are not identified.
For an open subset \(O\subseteq M_f\), write \(V_a=\{t:(a,t)\in q^{-1}(O)\}\) and \(V_b=\{t:(b,t)\in q^{-1}(O)\}\), interpreting the bottom points as representatives of \(y\). The quotient criterion says that \(V_a,V_b\) must be open in \(I\), and that \(y\in O\) exactly when both \(0\in V_a\) and \(0\in V_b\). In detail, if \(y\in O\), both bottom points must be included; if \(y\notin O\), neither can be included. There is no further matching condition at positive heights. This describes the topology directly from the quotient, including the neighborhoods of the common attached point.
The canonical target map sends \(y\) to the common bottom point. The canonical source map sends \(a\) and \(b\) to the two distinct top points represented by \((a,1)\) and \((b,1)\). Although \(f\) sends both source points to \(y\), \(i\) does not identify them.
The Canonical Copies of the Target and Source
Proof. The map \(j\) is injective because no two distinct points of \(Y\) are identified. The map \(i\) is injective because distinct points \((x,1)\) and \((x',1)\) are not identified. Both maps are continuous: \(j\) is the quotient map \(q\) composed with the inclusion of the \(Y\) component into the disjoint union, and \(i\) is \(q\) composed with the continuous map \(x\mapsto(x,1)\) into \(X\times I\).
These facts establish continuity of \(j\) and \(i\). To establish continuity of their inverse maps, we must show that images of open subsets of \(Y\) and \(X\) are open in the corresponding subspaces; we do this next.
Let \(U\subseteq Y\) be open. The subset $$ G_U=U\bigsqcup\bigl(f^{-1}(U)\times[0,1/2)\bigr) $$ is open in the disjoint union. Indeed, \(f^{-1}(U)\) is open by continuity of \(f\), and \([0,1/2)\) is open in the subspace \(I\). It is saturated: its bottom points are exactly \(f^{-1}(U)\times\{0\}\), matching the points of \(U\). Therefore \(q(G_U)\) is open in \(M_f\). Its intersection with \(j(Y)\) is exactly \(j(U)\), since it contains precisely the target points in \(U\), and no positive-height cylinder point is identified with a target point. Hence \(j(U)\) is open in the subspace \(j(Y)\).
Now let \(A\subseteq X\) be open. The set \(H_A=A\times(1/2,1]\) is open in \(X\times I\), and \(G= \varnothing\bigsqcup H_A\) is open in the disjoint union. It is saturated because it contains no bottom-cylinder point and no target point. Thus \(q(G)\) is open in \(M_f\), and \(q(G)\cap i(X)=i(A)\). Consequently \(i(A)\) is open in the subspace \(i(X)\). For an injective map, openness of the images of all open subsets is exactly continuity of the inverse from the image: the preimage under that inverse of \(j(U)\), or respectively \(i(A)\), is \(U\), or respectively \(A\). Both inverse maps are continuous, so both canonical maps are embeddings. These arguments also apply when either space is empty. \(\square\)
Worked Example: A Target Point Not Hit by the Map
Let \(X=\{a\}\) and \(Y=\{y,z\}\), both discrete, with \(f(a)=y\). The quotient consists of a copy of \(I\), whose point \(0\) represents both \(y\) and \((a,0)\), together with the separate point \(z\). To calculate the open sets, let \(V\subseteq I\) be the cylinder coordinates of a preimage and let \(U\subseteq Y\) be its target part. The preimage is open exactly when \(V\) is open in \(I\) and \(U\) is open in \(Y\); it is saturated exactly when \(y\in U\) if and only if \(0\in V\). Membership of \(z\) in \(U\) is unrestricted by the attachment.
Thus the quotient topology is precisely that of the disjoint union \(I\bigsqcup\{z\}\), with \(j(y)\) corresponding to \(0\), \(j(z)\) to the separate point, and \(i(a)\) to \(1\). For example, the image of the open interval \((1/2,1]\subseteq X\times I\) is open and meets the source image only at \(i(a)\), as the embedding proof predicts. The unattached target point remains part of the target copy; the construction does not require \(f\) to be surjective.
A Common Mistake and Why the Construction Matters
Common pitfall. The source inclusion is sometimes mistakenly defined by \(x\mapsto q(x,0)\). This is not the canonical source copy: it lands at \(j(f(x))\). In the two-point example above, this incorrect map sends both \(a\) and \(b\) to \(y\), so it is not injective. The canonical source map uses the opposite end, \(i(x)=q(x,1)\); its values at \(a\) and \(b\) are distinct. The distinction matters whenever \(f\) is not injective.
The quotient description also gives a practical test for proposed open sets: their preimages must be open in both pieces and must match exactly at the attached end. This is why the construction applies to arbitrary continuous maps without assuming that the map is one-to-one or onto. The theorem identifies two embedded subspaces inside the resulting space: the target sits at the attached end, and the source sits at the free end. It does not assert that either inclusion is an open map into the whole mapping cylinder.
| Construction feature | Exact description |
|---|---|
| Points identified | \((x,0)\) is identified with \(f(x)\) |
| Canonical target map | \(j(y)=q(y)\) |
| Canonical source map | \(i(x)=q(x,1)\) |
| Open set test | A subset is open exactly when its quotient preimage is open |
| Embedding conclusion | Both \(j\) and \(i\) are embeddings |
Check Your Understanding
Use the quotient definition and the endpoint matching condition in each question.
- For the constant map from a two-point discrete space to a one-point space, which points of the cylinder are identified with the target point?
- Why is \(x\mapsto q(x,0)\) not necessarily an injective map from the source into the mapping cylinder?
- For an open set \(U\subseteq Y\), describe a saturated open subset of the disjoint union whose image intersects \(j(Y)\) in \(j(U)\).
- In the example with \(X=\{a\}\), \(Y=\{y,z\}\), and \(f(a)=y\), why does the point \(z\) remain separate from the cylinder?
- Which part of the embedding proof shows that the inverse \(i(X)\to X\) is continuous?