Images and Preimages of Sets

In topology and analysis, continuity of $f: X \to Y$ is defined as: $f^{-1}[U]$ is open in $X$ for every open $U \subseteq Y$. The unconditional preimage laws proven above are exactly what make arbitrary unions and finite intersections of open sets pull back to open sets, which is the backbone of the definition of a topology and why continuous functions compose well. In measure theory and probability, a function is measurable precisely when preimages of measurable sets are measurable, and the $\sigma$-algebra axioms (closed under countable union, complement) transfer perfectly through $f^{-1}$ by the same laws, which is why random variables can be built by composing measurable functions. In algebra, the image $f[H]$ of a subgroup $H$ under a group homomorphism $f$ is again a subgroup, and the preimage $f^{-1}[K]$ of a subgroup $K$ is again a subgroup — both facts are proved by combining the definitions here with the homomorphism property, and the kernel $f^{-1}[\{e\}]$ is the cornerstone of the First Isomorphism Theorem. In cryptography and computer science, hash functions and encryption maps are studied through exactly these lenses: a hash function's collision resistance is a statement about how badly the intersection law for images fails (many inputs colliding to one output), while a cipher's invertibility is precisely the injectivity hypothesis needed for the round-trip equalities to hold. In statistics, the preimage of an event under a statistic (a function of data) is used to compute probabilities of observed outcomes, relying on the preimage-complement and preimage-union laws to manipulate events. Finally, this machinery directly foreshadows Finite Sets and Counting: counting arguments constantly ask "how big is $f[S]$?" (images can only shrink or preserve cardinality, never grow it, and equality of cardinality forces injectivity on $S$), and pigeonhole-style reasoning is essentially a cardinality refinement of the image–intersection failure exhibited above. Understanding images and preimages of sets is also a direct prerequisite for Cardinality of Sets, since injections and bijections are defined and compared using exactly the image/preimage vocabulary developed here, building on Composition of Functions to track how images transform under composed maps.

:::quiz[Let $f : \mathbb{Z} \to \mathbb{Z}$, $f(n) = n^2$. Let $S_1 = \{-2, 5\}$ and $S_2 = \{2, 7\}$. What is $f[S_1 \cap S_2]$?]

• $\{4\}$
• $\{4, 25, 49\}$
• $\varnothing$ ✓
• $\{4, 49\}$

:::quiz[Which single hypothesis on $f: A \to B$ guarantees $f[f^{-1}[T]] = T$ for every $T \subseteq B$?]

• $f$ is injective
• $f$ is surjective ✓
• $f$ is the identity function
• $A$ and $B$ are finite with $|A| = |B|$

:::quiz[Which of these correctly identifies the source of the failure of $f[S_1 \cap S_2] = f[S_1] \cap f[S_2]$ in general?]

• Unions of sets are not always well-defined
• Existential quantifiers cannot be swapped past each other
• Two distinct elements in $S_1 \setminus S_2$ and $S_2 \setminus S_1$ respectively can share the same image, without being in the intersection ✓
• Preimages are not defined when $f$ is not injective

:::faq[Is it true that $f[A \setminus S] = B \setminus f[S]$ when $f$ is a bijection?] Yes, but only under the full bijection hypothesis (both injective and surjective), and it is worth proving rather than assuming. If $f$ is a bijection, $f^{-1}$ is an honest inverse function, and $f[A\setminus S] = \{f(x) : x \notin S\}$. Injectivity ensures no element of $f[S]$ can also arise from a point outside $S$, and surjectivity ensures every point of $B \setminus f[S]$ is hit by some $x$, which (by injectivity again) must lie outside $S$. Both hypotheses are used essentially; the counterexample block above shows the identity fails outright once injectivity alone is dropped, and a similar example shows surjectivity alone is not sufficient either.

:::faq[Does $f[S_1] \subseteq f[S_2]$ follow from $S_1 \subseteq S_2$?] Yes, and this monotonicity needs no hypothesis on $f$. If $y \in f[S_1]$, there is $x \in S_1$ with $f(x) = y$; since $S_1 \subseteq S_2$, also $x \in S_2$, so $y \in f[S_2]$. The identical argument shows $f^{-1}[T_1] \subseteq f^{-1}[T_2]$ whenever $T_1 \subseteq T_2$. Monotonicity is actually the easy half of the union and intersection laws: the union law's "$\subseteq$" direction and the intersection law's inclusion (item 5) both follow immediately from monotonicity applied to each $S_i \subseteq \bigcup_j S_j$ or $\bigcap_j S_j \subseteq S_i$ respectively.

:::faq[Can the fibers $f^{-1}[\{y\}]$ for $y$ ranging over $B$ be used to reconstruct $A$?] Yes: the collection of nonempty fibers $\{f^{-1}[\{y\}] : y \in \operatorname{ran}(f)\}$ partitions $A$ into disjoint, nonempty blocks, one per value actually achieved. This follows directly from the preimage laws: fibers over distinct points are disjoint because $f^{-1}[\{y_1\}] \cap f^{-1}[\{y_2\}] = f^{-1}[\{y_1\} \cap \{y_2\}] = f^{-1}[\varnothing] = \varnothing$ when $y_1 \ne y_2$, using law (2). Their union is all of $A$ because every $x \in A$ lies in the fiber over $f(x)$. This partition-by-fibers viewpoint is exactly the bridge used later to compare $|A|$ and $|\operatorname{ran}(f)|$ in counting arguments.

Ask the tutor: Why does the preimage complement law hold with no hypotheses, while no version of a complement law holds for direct images even under injectivity alone?

Think of $f$ as a machine with input bin $A$ and output bin $B$. The direct image $f[S]$ answers: "if I feed in exactly the elements of $S$, what collection of outputs do I get?" It is a forward-looking, existential notion — an output $y$ belongs to $f[S]$ if some element of $S$ produces it. This existential quantifier is precisely the source of all the trouble with intersections: two different elements $x_1 \in S_1$ and $x_2 \in S_2$ can both map to the same $y$, so $y$ can appear in both $f[S_1]$ and $f[S_2]$ even though $x_1, x_2$ need not both lie in $S_1 \cap S_2$. The preimage $f^{-1}[T]$ answers a different, backward-looking question: "which inputs are guaranteed to land in $T$?" This is a universal, pointwise membership test: $x \in f^{-1}[T]$ exactly when the single, well-defined value $f(x)$ satisfies $f(x) \in T$. Because each $x$ has exactly one output $f(x)$, testing membership in $T$ is a condition on $x$ alone, with no existential ambiguity — this is why preimages commute with all Boolean operations without any hypothesis on $f$.

Example 1 (a genuinely non-injective, non-surjective map, tracking all six operations). Let $A = \{1,2,3,4,5\}$ and $B = \{a,b,c,d,e,f\}$, and define $f: A \to B$ by

So $\operatorname{ran}(f) = \{a,b,c\}$; $f$ is neither injective (e.g. $f(2)=f(3)$) nor surjective (e.g. $d,e,f \notin \operatorname{ran}(f)$). Take $S_1 = \{1,2,3\}$ and $S_2 = \{2,4,5\}$. Images. $f[S_1] = \{f(1),f(2),f(3)\} = \{a,b\}$. $f[S_2] = \{f(2),f(4),f(5)\} = \{b,c,a\} = \{a,b,c\}$. Union law (item 4). $S_1 \cup S_2 = \{1,2,3,4,5\} = A$, so $f[S_1 \cup S_2] = f[A] = \{a,b,c\}$. On the other hand $f[S_1] \cup f[S_2] = \{a,b\} \cup \{a,b,c\} = \{a,b,c\}$. These agree, confirming (4). Intersection law (item 5), strict inclusion this time. $S_1 \cap S_2 = \{2\}$, so $f[S_1 \cap S_2] = \{f(2)\} = \{b\}$. But $f[S_1] \cap f[S_2] = \{a,b\} \cap \{a,b,c\} = \{a,b\}$. Indeed

a strict inclusion, exactly as predicted since $f$ is not injective: the element $a$ appears in both $f[S_1]$ and $f[S_2]$ via different preimages ($1 \in S_1$ and $5 \in S_2$, with $f(1)=f(5)=a$, but $1 \notin S_2$ and $5 \notin S_1$, so $1,5 \notin S_1 \cap S_2$). Round trip forward (item 7). Take $S = \{2\}$. Then $f[S] = \{b\}$, and $f^{-1}[f[S]] = f^{-1}[\{b\}] = \{2,3\}$ (both 2 and 3 map to $b$). Indeed $S = \{2\} \subsetneq \{2,3\} = f^{-1}[f[S]]$, strict because $f$ is not injective and $3$ "rides along" with $2$. Round trip backward (item 8). Take $T = \{b, d\}$ ($d \notin \operatorname{ran}(f)$). Then $f^{-1}[T] = \{x : f(x) \in \{b,d\}\} = \{2,3\}$ (nothing maps to $d$). So $f[f^{-1}[T]] = f[\{2,3\}] = \{b\}$. Indeed $\{b\} \subsetneq \{b,d\} = T$, strict because $f$ is not surjective ($d$ is never hit, so it can never re-enter the image after being "filtered" by the preimage). Preimage complement law (item 3), a direct check. Let $T = \{a,c\}$, so $B \setminus T = \{b,d,e,f\}$. We compute $f^{-1}[T] = \{x : f(x) \in \{a,c\}\} = \{1,4,5\}$, so $A \setminus f^{-1}[T] = \{2,3\}$. Directly, $f^{-1}[B\setminus T] = \{x : f(x) \in \{b,d,e,f\}\} = \{2,3\}$ (since only $2,3 \mapsto b$, and nothing maps to $d,e,f$). Both computations give $\{2,3\}$, confirming (3) concretely.

Example 2 (an infinite setting: $f: \mathbb{Z} \to \mathbb{Z}$, $f(n) = n^2$, illustrating the theorem with infinite index families). Let $f : \mathbb{Z} \to \mathbb{Z}$, $f(n) = n^2$. This $f$ is neither injective (since $f(-2) = f(2) = 4$) nor surjective (negative integers and non-squares are missed). For each $k \ge 1$ let $S_k = \{-k, k\}$ (so $f[S_k] = \{k^2\}$), and consider the family $\{S_k\}_{k=1}^{\infty}$. Union law. $\bigcup_{k=1}^\infty S_k = \mathbb{Z} \setminus \{0\}$. Then

and $\bigcup_{k=1}^\infty f[S_k] = \bigcup_{k=1}^\infty \{k^2\} = \{k^2 : k \ge 1\}$. These match exactly, confirming (4) in an infinite setting. Intersection failing, infinite version. Let $S_1 = \{-1, 3\}$ and $S_2 = \{1, 3\}$ (asymmetric, sharing only the point $3$ in a nontrivial way but also each containing a point whose image collides with the other's). Then $S_1 \cap S_2 = \{3\}$, so $f[S_1 \cap S_2] = \{9\}$. But $f[S_1] = \{1, 9\}$ and $f[S_2] = \{1,9\}$, so $f[S_1] \cap f[S_2] = \{1,9\}$. Indeed

because $-1 \in S_1$ and $1 \in S_2$ collide under $f$ (both square to $1$) without $-1$ or $1$ lying in the intersection $S_1 \cap S_2$ — the mechanism from the general proof of item (5), realized concretely. Preimage of an intersection, done correctly (contrast). Let $T_1 = \{0,1,4,9,16,\dots\}$ (all squares, including $0$) and $T_2 = \{n \in \mathbb{Z} : n \ge 5\}$. Then $T_1 \cap T_2 = \{9, 16, 25, \dots\}$ (squares that are at least $5$), so

On the other hand $f^{-1}[T_1] = \mathbb{Z}$ (every integer squares to something in $T_1$, since $T_1$ contains all perfect squares) and $f^{-1}[T_2] = \{n : n^2 \ge 5\} = \{n : |n| \ge 3\}$ (since $2^2 = 4 < 5$ but $3^2 = 9 \ge 5$). So

which matches $f^{-1}[T_1 \cap T_2]$ exactly, confirming (2) holds with no injectivity assumption whatsoever — even though this very same $f$ badly violates the analogous law for images.

:::warning[The classic error: assuming $f[S_1 \cap S_2] = f[S_1] \cap f[S_2]$ in general] The single most common mistake is to treat images as if they distribute over intersection the same way preimages do, silently assuming injectivity without stating it. The correct universally valid statement is only the inclusion $f[S_1 \cap S_2] \subseteq f[S_1] \cap f[S_2]$; equality requires injectivity and must be justified, never asserted "by symmetry" with the (unconditional) preimage law. A related quantifier trap: "$f$ injective on $S_1 \cup S_2$" is what is really needed locally, but it suffices to assume $f$ injective on all of $A$, which is the clean global hypothesis stated in the theorem.