Fibers Record Which Points Are Identified
A surjection can identify distinct points of its domain by sending them to the same point in its codomain. The set of all points sent to a particular value is its fiber. A subset of the domain respects those identifications precisely when it includes a whole fiber whenever it includes any point of that fiber. Such subsets are called saturated.
Let \(X\) and \(Y\) be sets and let \(f:X\to Y\) be a surjection. When topologies are relevant, assume \(X\) and \(Y\) have specified topologies; the definitions in this tutorial depend only on the underlying sets and the map. For \(y\in Y\), the fiber over \(y\) is the preimage \(f^{-1}(\{y\})\). Surjectivity means each such fiber is nonempty. Equivalently, the relation \(x\sim x'\) defined by \(f(x)=f(x')\) divides \(X\) into equivalence classes, and those classes are exactly the fibers.
The condition is local to each fiber: a saturated set cannot include only some of the points that the map identifies. The topology on \(X\) or \(Y\) does not affect whether a set is saturated. In particular, being open or closed is a separate question.
A Test for Saturation
Proof. First, \(A\subseteq f^{-1}(f(A))\): if \(x\in A\), then \(f(x)\in f(A)\), so \(x\in f^{-1}(f(A))\). Now suppose \(A\) is saturated. If \(x\in f^{-1}(f(A))\), then \(f(x)\in f(A)\), so there is some \(a\in A\) with \(f(x)=f(a)\). Saturation gives \(x\in A\). Thus \(f^{-1}(f(A))\subseteq A\), proving equality.
Conversely, suppose \(f^{-1}(f(A))=A\). If \(a\in A\) and \(f(x)=f(a)\), then \(f(x)\in f(A)\), so \(x\in f^{-1}(f(A))=A\). Hence \(A\) contains every fiber it meets and is saturated.
For the final claim, set \(S=f^{-1}(f(A))\). The equality just proved, applied to \(S\), gives \(f^{-1}(f(S))=S\), because \(f(S)=f(A)\). Thus \(S\) is saturated. It contains \(A\), as shown above. If \(C\) is any saturated set containing \(A\), then \(f(A)\subseteq f(C)\), so \(f^{-1}(f(A))\subseteq f^{-1}(f(C))=C\). Therefore \(S\subseteq C\), proving it is the smallest such set. If \(A=\varnothing\), then \(f(A)=\varnothing\) and \(S=\varnothing\), so the statement includes the empty case. \(\square\)
The set \(f^{-1}(f(A))\) is often called the saturation of \(A\). To calculate it, first find all values \(f(a)\) for \(a\in A\), then include every point of \(X\) mapping to one of those values.
Worked Example: The Square Map
Let \(f:\mathbb{R}\to[0,\infty)\) be \(f(x)=x^2\), with both spaces carrying their standard subspace or Euclidean topologies. This is a surjection: for each \(y\ge0\), the real number \(\sqrt y\) satisfies \(f(\sqrt y)=y\). The fiber over \(y=0\) is \(\{0\}\); for \(y>0\), it is \(\{-\sqrt y,\sqrt y\}\), since \(x^2=y\) is equivalent to \(x=\pm\sqrt y\).
Take \(A=[1,2]\). Its image is \(f(A)=[1,4]\), since squaring is increasing on this interval and sends its endpoints to \(1\) and \(4\). The saturation is $$ f^{-1}([1,4])=[-2,-1]\cup[1,2]. $$ Indeed, \(x^2\in[1,4]\) means \(1\le x^2\le4\), which is equivalent to \(1\le |x|\le2\), or \(-2\le x\le-1\) or \(1\le x\le2\). The endpoints are included because the image interval includes \(1\) and \(4\). Thus \(A\) is not saturated: it contains \(1\), but not the other point \(-1\) in the same fiber.
Check: the image of \([-2,-1]\cup[1,2]\) is \([1,4]\), and its preimage is again \([-2,-1]\cup[1,2]\). So the computed saturation is saturated. More generally, for this map a set is saturated exactly when it is symmetric under \(x\mapsto -x\): every nonzero fiber consists of a pair \(\{x,-x\}\), and the fiber at zero is already a singleton.
Worked Example: A Projection and Vertical Fibers
Give \(\mathbb{R}^2\) and \(\mathbb{R}\) their standard topologies, and let \(\pi_1:\mathbb{R}^2\to\mathbb{R}\) be the projection \(\pi_1(x,y)=x\). It is surjective because \(\pi_1(x,0)=x\) for every real \(x\). For each \(c\in\mathbb{R}\), its fiber is the full vertical line \(\{c\}\times\mathbb{R}\): a point \((x,y)\) maps to \(c\) exactly when \(x=c\).
Let \(A=([0,1]\times\{0\})\cup(\{2\}\times\{3\})\). The image is \(\pi_1(A)=[0,1]\cup\{2\}\). In this expression, both endpoints \(0\) and \(1\) belong to \([0,1]\), and \(2\) belongs as a separate point. Therefore $$ \pi_1^{-1}(\pi_1(A))=([0,1]\times\mathbb{R})\cup(\{2\}\times\mathbb{R}). $$ This set is the saturation of \(A\). The original \(A\) is not saturated: it contains \((0,0)\), for example, but does not contain \((0,1)\), even though both points have first coordinate \(0\).
Check: every vertical line with first coordinate in \([0,1]\cup\{2\}\) is included in the displayed set, and no other vertical line meets it. Hence the set is a union of complete fibers. In general, a subset of \(\mathbb{R}^2\) is saturated for \(\pi_1\) exactly when it is a union of full vertical lines.
Saturated Sets as Preimages
Proof. Suppose \(A=f^{-1}(B)\). If \(x\in A\) and \(f(x')=f(x)\), then \(f(x)\in B\), so \(f(x')\in B\) and \(x'\in A\). Thus \(A\) is saturated.
Conversely, suppose \(A\) is saturated and put \(B=f(A)\). The theorem gives \(A=f^{-1}(f(A))=f^{-1}(B)\). To prove uniqueness, suppose \(f^{-1}(B)=f^{-1}(D)\) for subsets \(B,D\subseteq Y\). If \(y\in B\), surjectivity provides \(x\in X\) with \(f(x)=y\). Then \(x\in f^{-1}(B)=f^{-1}(D)\), so \(y=f(x)\in D\). This proves \(B\subseteq D\). The same argument with \(B,D\) exchanged gives \(D\subseteq B\), hence \(B=D\). If \(Y=\varnothing\), surjectivity forces \(X=\varnothing\), and the only possible sets are both empty; the conclusion still holds. \(\square\)
Surjectivity matters for uniqueness: if \(f\) misses points of \(Y\), changing \(B\) only at those points does not change \(f^{-1}(B)\). The existence direction still works for any map, by taking \(B=f(A)\), but the one-to-one correspondence between subsets of \(Y\) and saturated subsets of \(X\) uses surjectivity.
Worked Example: Identifying the Endpoints of an Interval
Let \(I=[0,1]\) have its subspace topology from \(\mathbb{R}\), let \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\) have its subspace topology, and define \(q:I\to S^1\) by \(q(t)=(\cos(2\pi t),\sin(2\pi t))\). This map is surjective because every point of the unit circle has an angle between \(0\) and \(2\pi\). Its fibers are \(\{0,1\}\) over \((1,0)\), and singleton sets \(\{t\}\) for \(0<t<1\). Indeed, the parametrization repeats after an integer change in \(t\), and two parameters in \([0,1]\) differ by an integer only when they are equal or are the two endpoints.
Take \(A=[0,\tfrac12]\). Its image consists of the circle points parametrized by \(0\le t\le\tfrac12\). This includes \((1,0)\), whose entire fiber is \(\{0,1\}\), and the fibers of all other included points are singletons. Therefore $$ q^{-1}(q(A))=[0,\tfrac12]\cup\{1\}. $$ The endpoint \(1\) is included in the saturation because \(q(1)=(\cos(2\pi),\sin(2\pi))=(1,0)=q(0)\). The original \(A\) is not saturated, since it contains \(0\) but omits \(1\).
Check: the displayed set contains both \(0\) and \(1\), and every other point it contains has a singleton fiber contained in the set. It is therefore saturated. In contrast, \([\tfrac14,\tfrac34]\) is saturated: it contains neither endpoint, and every fiber it meets is a singleton.
A Common Mistake and Why Saturation Matters
Common pitfall. A subset is not saturated just because it is the image of some subset, or because it contains several points from each fiber. For the square map, the incorrect calculation \(f^{-1}(f([1,2]))=[1,2]\) keeps only the positive square roots. But \(f([1,2])=[1,4]\), and the full preimage also contains every negative square root: \(f^{-1}([1,4])=[-2,-1]\cup[1,2]\). The correction is to take the preimage of the entire image, not merely to retain the original points.
Saturated sets are the subsets of a domain that can be described entirely in terms of the map’s output: each has the form \(f^{-1}(B)\) for a subset \(B\) of the codomain. This is why they are central when a construction treats points in the same fiber as one identified point. The key test remains set-theoretic: include a fiber completely or do not include it at all. Whether the resulting set is open, closed, or has another topological property requires a separate argument.
Check Your Understanding
For each question, identify the relevant fibers before deciding whether the set is saturated.
- For \(f:\mathbb{R}\to[0,\infty)\), \(f(x)=x^2\), find the saturation of \([-3,-2]\).
- For the projection \(\pi_1:\mathbb{R}^2\to\mathbb{R}\), is \(\mathbb{R}\times\{0\}\) saturated? Explain using its fibers.
- For the interval map \(q:I\to S^1\) above, determine whether \([0,\tfrac13]\cup\{1\}\) is saturated.
- Prove directly that the preimage of any subset of the codomain is saturated under a surjection.
- Give an example of a subset \(A\) that is not saturated, and compute \(f^{-1}(f(A))\) for your chosen map and set.