Why Inverse Images Have Reliable Set Rules
Given a function \(f:X\to Y\), a subset of the codomain determines a subset of the domain: the points whose images land in that subset. This operation is useful in topology because it translates conditions on points in \(Y\) into conditions on points in \(X\). Its basic rules are set-theoretic; they hold whether or not \(f\) is continuous.
Throughout, \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) are topological spaces, and \(f:X\to Y\) is a function. For \(A\subseteq Y\), define the inverse image of \(A\) by \(f^{-1}(A)=\{x\in X:f(x)\in A\}\). The inverse image is a subset of \(X\), not of \(Y\). In particular, complements of inverse images are taken in \(X\), while complements of subsets of \(Y\) are taken in \(Y\).
Proof. First suppose \(J\) is nonempty. A point \(x\in X\) belongs to \(f^{-1}(\bigcup_{i\in J}A_i)\) exactly when \(f(x)\in A_i\) for at least one \(i\in J\). This holds exactly when \(x\in f^{-1}(A_i)\) for at least one \(i\in J\), which is membership in \(\bigcup_{i\in J}f^{-1}(A_i)\). For intersections, \(x\in f^{-1}(\bigcap_{i\in J}A_i)\) exactly when \(f(x)\in A_i\) for every \(i\in J\). This holds exactly when \(x\in f^{-1}(A_i)\) for every \(i\in J\), which is membership in \(\bigcap_{i\in J}f^{-1}(A_i)\).
For the empty family, the union is \(\varnothing\) and the intersection is \(Y\). Their inverse images are respectively \(\varnothing\) and \(X\), which agree with the empty union and empty intersection of subsets of \(X\). For the complement identity, \(x\in f^{-1}(Y\setminus A)\) exactly when \(f(x)\notin A\), and this is equivalent to \(x\notin f^{-1}(A)\). Finally, no point maps into \(\varnothing\), and every point maps into \(Y\), so \(f^{-1}(\varnothing)=\varnothing\) and \(f^{-1}(Y)=X\). \(\square\)
The proof uses only the definition of inverse image and membership in sets. There is no finiteness restriction: both identities for unions and intersections hold for infinite families as well as finite ones. Complements, however, must be interpreted in the appropriate spaces: the complement of \(A\) is in \(Y\), while the complement of \(f^{-1}(A)\) is in \(X\).
Using the Identities to Calculate
Worked Example: A Squaring Function
Give \(\mathbb{R}\) its standard topology and define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x^2\). Let \(A=[0,4]\) and \(B=[1,9]\). We calculate the preimages of their union, intersection, and a complement.
First, \(x\in f^{-1}(A)\) exactly when \(0\le x^2\le4\), so \(-2\le x\le2\) and \(f^{-1}(A)=[-2,2]\). Likewise, \(x\in f^{-1}(B)\) exactly when \(1\le x^2\le9\), so \(x\in[-3,-1]\cup[1,3]\). Thus \(f^{-1}(A)\cup f^{-1}(B)=[-3,3]\): the two intervals overlap on \([-2,-1]\) and \([1,2]\), and together cover all points from \(-3\) to \(3\).
Since \(A\cup B=[0,9]\), its preimage consists of the \(x\) satisfying \(0\le x^2\le9\), namely \([-3,3]\). This verifies the union identity in this calculation. Also \(A\cap B=[1,4]\). The condition \(1\le x^2\le4\) gives \(x\in[-2,-1]\cup[1,2]\), which is exactly \(f^{-1}(A)\cap f^{-1}(B)\).
For the complement, \(Y\setminus A=(-\infty,0)\cup(4,\infty)\). No square is negative, and \(x^2>4\) exactly when \(x<-2\) or \(x>2\). Therefore \(f^{-1}(Y\setminus A)=(-\infty,-2)\cup(2,\infty)\). On the other hand, \(X\setminus f^{-1}(A)=\mathbb{R}\setminus[-2,2]=(-\infty,-2)\cup(2,\infty)\), as required. The strict endpoints matter: \(x=\pm2\) maps to \(4\), which is in \(A\), not in its complement.
Worked Example: A Map That Misses a Codomain Point
Let \(X=\{p,q,r\}\) and \(Y=\{0,1,2\}\), both with the discrete topology, in which every subset is open. Define \(f:X\to Y\) by \(f(p)=0\), \(f(q)=1\), and \(f(r)=1\). Take \(A=\{0,2\}\) and \(B=\{1,2\}\).
By checking the image of each point, \(f^{-1}(A)=\{p\}\) and \(f^{-1}(B)=\{q,r\}\). Their union is \(X\). Since \(A\cup B=Y\), the identity gives \(f^{-1}(A\cup B)=f^{-1}(Y)=X\), matching that union. Their intersection is empty, since \(A\cap B=\{2\}\) and no point maps to \(2\). Thus \(f^{-1}(A\cap B)=\varnothing=f^{-1}(A)\cap f^{-1}(B)\).
The complement calculation also shows the ambient sets at work. Here \(Y\setminus A=\{1\}\), so \(f^{-1}(Y\setminus A)=\{q,r\}\). Meanwhile \(X\setminus f^{-1}(A)=X\setminus\{p\}=\{q,r\}\). Finally, \(f^{-1}(\{0,1\})=X=f^{-1}(Y)\), even though \(\{0,1\}\ne Y\). The equality occurs because \(f\) never takes the value \(2\); inverse images cannot detect differences between subsets of \(Y\) that occur only at missed values.
What Preimages Can Tell Us About Subsets
Proof. Suppose first that \(f^{-1}(A)\subseteq f^{-1}(B)\). If \(y\in f(X)\cap A\), then \(y=f(x)\) for some \(x\in X\), and \(y\in A\) gives \(x\in f^{-1}(A)\). The assumed inclusion gives \(x\in f^{-1}(B)\), so \(y=f(x)\in B\). Hence \(f(X)\cap A\subseteq B\). Conversely, suppose \(f(X)\cap A\subseteq B\), and let \(x\in f^{-1}(A)\). Then \(f(x)\in f(X)\cap A\), so \(f(x)\in B\). Therefore \(x\in f^{-1}(B)\). This proves the first equivalence.
For the equality statement, \(f^{-1}(A)=f^{-1}(B)\) holds exactly when both \(f^{-1}(A)\subseteq f^{-1}(B)\) and \(f^{-1}(B)\subseteq f^{-1}(A)\) hold. By the first part, these inclusions say respectively that \(f(X)\cap A\subseteq B\) and \(f(X)\cap B\subseteq A\). Together they say that membership in \(A\) and \(B\) agrees at every point of \(f(X)\). Conversely, that agreement gives both inclusions and hence equality of the preimages. If \(f\) is surjective, then \(f(X)=Y\), so this equality condition says \(A=B\). \(\square\)
This result separates two questions that are sometimes confused. A set inclusion \(A\subseteq B\) always gives an inclusion of preimages. The converse need not hold unless the map reaches all the points relevant to the comparison. The finite example above exhibits this limitation: the two sets \(\{0,1\}\) and \(Y\) have the same preimage because they agree on the image \(\{0,1\}\) of \(f\).
Worked Example: Inverse Images and Openness
Give \(X=\mathbb{R}\) the standard topology and \(Y=\{a,b\}\) the discrete topology. Define \(f:X\to Y\) by \(f(x)=a\) when \(x<0\), and \(f(x)=b\) when \(x\ge0\). The open subsets of \(Y\) are \(\varnothing,\{a\},\{b\},Y\). Their inverse images are, respectively, \(\varnothing,(-\infty,0),[0,\infty),\mathbb{R}\).
The sets \(\varnothing\) and \(\mathbb{R}\) are open, and \((-\infty,0)\) is open: for any \(x<0\), the interval \((x-|x|/2,x+|x|/2)\) lies below \(0\) and contains \(x\). But \([0,\infty)\) is not open. If it were open, it would contain an open interval around \(0\); every such interval contains negative points, which are not in \([0,\infty)\). Thus this particular map does not send the preimage of every open subset of \(Y\) to an open subset of \(X\).
The inverse-image identities still hold. For example, \(\{a\}\cup\{b\}=Y\), and \(f^{-1}(\{a\})\cup f^{-1}(\{b\})=(-\infty,0)\cup[0,\infty)=\mathbb{R}=f^{-1}(Y)\). The identity is valid even though one of these two preimages is not open. In the standard definition, a continuous map is a map for which the preimage of every open subset of the codomain is open in the domain. The identities calculate preimages; continuity concerns the additional topological property of those preimages.
A Common Mistake: Swapping Union and Intersection
Common pitfall. The union identity can be mistakenly written with an intersection on the right. In the finite example, take \(A=\{0\}\) and \(B=\{1\}\). Then \(f^{-1}(A)=\{p\}\) and \(f^{-1}(B)=\{q,r\}\). The incorrect calculation would give \(f^{-1}(A\cup B)=f^{-1}(A)\cap f^{-1}(B)=\varnothing\). But \(A\cup B=\{0,1\}\), whose preimage is \(\{p,q,r\}=X\), not the empty set. The correction is \(f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)=\{p\}\cup\{q,r\}=X\). A point maps into a union when its image belongs to at least one of the sets; it maps into an intersection only when its image belongs to every set.
The same membership-based reasoning is the safest way to use these rules. Before taking a complement, identify the ambient space. Before reversing an inclusion, check whether the map reaches all the relevant codomain points. And when applying the identities to open sets, keep separate the universal set-theoretic calculation from the question of whether the resulting subset of \(X\) is open.
Check Your Understanding
For each question, identify the ambient space in which each complement is taken.
- For an arbitrary function \(f:X\to Y\), prove directly from membership that \(f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)\).
- Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\). Find \(f^{-1}((-1,4])\) and \(f^{-1}(\mathbb{R}\setminus(-1,4])\).
- Give a function \(f:X\to Y\) and distinct subsets \(A,B\subseteq Y\) for which \(f^{-1}(A)=f^{-1}(B)\). Explain which codomain points the map misses.
- For a family \((A_i)_{i\in J}\), explain why the intersection identity remains valid when \(J\) is empty.
- In the third worked example, which open subset of \(Y\) has a preimage that is not open in \(X\)?