nLab interactions of images and pre-images with unions and intersections

Introduction

There is a battery of little set-theoretic lemmas which inevitably recur when laying the foundations of core mathematics, as when teaching the standard curriculum that includes basic topology, real analysis, algebra, and so forth. In this article we collect some of those core results having to do with the properties of image and pre-image operators, especially regarding their preservation (or not) of set-theoretic operations like unions and intersections.

The attitude and approach of the mathematics professor toward such routine bread-and-butter issues is a matter of some interest. A professional, upon being presented with any one of these “naive set theory” propositions, will probably be able to verify it on the spot using ordinary follow-your-nose logic, driven by the definitions. While being able to produce such verifications is part of basic training in mathematics, it is not quite the same as giving an explanation, and in fact even good mathematicians trained this way may struggle to remember “now which is it that direct images preserve, unions or intersections?”¹

A strong pedagogy would not only instill this sort of basic training, but make the battery of routine results more memorable by concentrating their essence in one or two basic principles that provide an explanatory basis for the rest. According to Lawvere 69, Lawvere 05, logic is an interlocking system of certain types of adjoint functors, and we believe putting those adjoint relationships front and center and seeing how the rest flows from them is an effective way to arrange the facts.

Statements

Proposition

(images preserve unions but not in general intersections)

Let $f \colon X \longrightarrow Y$ be a function between sets. Let $\{S_i \subset X\}_{i \in I}$ be an indexed set of subsets of $X$. Then

1. $im_f\left( \underset{i \in I}{\cup} S_i\right) = \left(\underset{i \in I}{\cup} im_f(S_i)\right)$ (the image under $f$ of a union of subsets is the union of the images)

2. $im_f\left( \underset{i \in I}{\cap} S_i\right) \subset \left(\underset{i \in I}{\cap} im_f(S_i)\right)$ (the image under $f$ of the intersection of the subsets is contained in the intersection of the images).

The injection in the second item is in general proper. If $f$ is an injective function then this is a bijection:

• $(f\,\text{injective}) \Rightarrow \left(im_f\left( \underset{i \in I}{\cap} S_i\right) = \left(\underset{i \in I}{\cap} im_f(S_i)\right)\right)$

provided that $I$ is inhabited.

Proposition

(pre-images preserve unions and intersections)

Let $f \colon X \longrightarrow Y$ be a function between sets. Let $\{T_i \subset Y\}_{i \in I}$ be an indexed set of subsets of $Y$. Then

1. $f^{-1}\left( \underset{i \in I}{\cup} T_i\right) = \left(\underset{i \in I}{\cup} f^{-1}(T_i)\right)$ (the pre-image under $f$ of a union of subsets is the union of the pre-images),

2. $f^{-1}\left( \underset{i \in I}{\cap} T_i\right) = \left(\underset{i \in I}{\cap} f^{-1}(T_i)\right)$ (the pre-image under $f$ of the intersection of the subsets is the intersection of the pre-images).

Proposition

(projection formula) Let $f: X \longrightarrow Y$ be a function between sets. Let $S \subset X$ be a subset of $X$, and $T \subseteq Y$ be a subset of $Y$. Then

$$f(S) \cap T = f(S \cap f^{-1}(T)).$$

Proofs via adjoints

The properties 1., 2. of Proposition may be proved by appeal to fundamental relationships between direct image and inverse image and the like, which category theorists call adjunctions (similar in form to adjoints in linear algebra). The advantage of this type of proof is that, despite its utter simplicity, it generalizes to much wider contexts (beyond elementary classical set theory).

Direct images preserve unions, inverse images preserve intersections

The first such relationship is:

Proposition

(preimage is right adjoint to image)
For all subsets $S \subseteq X$ and $T \subseteq Y$, we have

$$f(S) \subseteq T \qquad iff \qquad S \subseteq f^{-1}(T) \,.$$

One says in this situation that the direct image mapping $f(-)$ is left adjoint to the inverse image mapping $f^{-1}(-)$ (with respect to the posets of subsets, regarded as (0,1)-categories), or that inverse image is right adjoint to direct image. This terminology is guided by the formally similar usage in linear algebra where linear mappings $A$ and $B$ are adjoint if one has an equation

$$\langle A x, y \rangle = \langle x, B y \rangle$$

for all $x, y$ in suitable inner product spaces (and in fact the analogy is not idle; see for instance Baez 1997).

In that case, we have the following elementary inferences:

$$\array{ \bigcup_{i \in I} f(S_i) \subseteq T & iff & (\forall_{i \in I})\; f(S_i) \subseteq T \\ & iff & (\forall_{i \in I})\; S_i \subseteq f^{-1}(T) \\ & iff & \bigcup_{i \in I} S_i \subseteq f^{-1}(T) \\ & iff & f\left(\bigcup_{i \in I} S_i\right) \subseteq T }$$

where the first and third lines use the defining property of unions. Then, putting $T = \bigcup_{i \in I} f(S_i)$ and reasoning forward, we get $f\left(\bigcup_{i \in I} S_i\right) \subseteq \bigcup_{i \in I} f(S_i)$. On the other hand, putting $T = f\left(\bigcup_{i \in I} S_i\right)$ and reasoning backward, we get $\bigcup_{i \in I} f(S_i) \subseteq f\left(\bigcup_{i \in I} S_i\right)$. These two inclusions together give $\bigcup_{i \in I} f(S_i) = f\left(\bigcup_{i \in I} S_i\right)$.

The dual of this proof immediately gives the fact that inverse images preserve arbitrary intersections, but let us spell it out directly:

$$\array{ S \subseteq \bigcap_{i \in I} f^{-1}(T_i) & iff & (\forall_{i \in I})\; S \subseteq f^{-1}(T_i) \\ & iff & (\forall_{i \in I})\; f(S) \subseteq T_i \\ & iff & f(S) \subseteq \bigcap_{i \in I} T_i \\ & iff & S \subseteq f^{-1}\left(\bigcap_{i \in I} T_i \right) }$$

and again by using the reasoning forward/backward trick, we infer $\bigcap_{i \in I} f^{-1}(T_i) = f^{-1}\left(\bigcap_{i \in I} T_i\right)$.

Remark

The “reasoning forward/backward trick” may be summarized by saying that in a poset, we have $A = B$ iff ($A \leq T \Leftrightarrow B \leq T$), or dually that $A = B$ iff ($S \leq A \Leftrightarrow S \leq B$). This trick is vastly extrapolated by the Yoneda lemma.

Remark

People who work with categorial forms of logic denote direct images $f(S)$ by $\exists_f(S)$. The suggestion is to view existential quantification as corresponding to taking of a direct image. For example, given a property $P$ of pairs $(x, y)$, i.e. a subset $P \subseteq X \times Y$ where we write $P(x, y)$ to say $(x, y) \in P$ holds, the set of $y$ such that $(\exists_{x \in X})\; P(x, y)$ holds is exactly the direct image $\pi_2(P)$ under the projection map $\pi_2: X \times Y \to Y$. This suggests furthermore a useful extended meaning of existential quantification, by considering direct images along more general maps, not just projection maps. With this in mind, category theorists often denote $f(S) = \{y \in Y: (\exists_{x: f(x) = y})\; x \in S\}$ by $\exists_f(S)$, giving an operator $\exists_f: P(X) \to P(Y)$ between power sets. This is left adjoint to the taking of inverse images $T \mapsto f^{-1}(T)$ as an operator $f^{-1}: P(Y) \to P(X)$, also denoted by $f^\ast: P(Y) \to P(X)$. The adjointness relationship between direct image and inverse image, denoted by $\exists_f \dashv f^\ast$, is an example of a famous slogan due to Lawvere: “Logical quantification is adjoint to substitution” (with resonances far beyond the purview of logic as ordinarily conceived; see Remark ).

As for direct images and intersections, we have

$$f\left(\bigcap_{i \in I} S_i\right) \subseteq \bigcap_{i \in I} f(S_i)$$

because this is equivalent to $(\forall_{i \in I})\; f\left(\bigcap_{i \in I} S_i\right) \subseteq f(S_i)$, which is obvious because for all $i \in I$, we have $\bigcap_{i \in I} S_i \subseteq S_i$, and $f(A) \subseteq f(B)$ if $A \subseteq B$.

This inclusion is not usually an equality. For example, in the cartesian space $\mathbb{R}^2$ with $x y$-coordinates, the projection $(x, y) \mapsto x$ does not preserve the intersection of the lines $y = 1$ and $y = 2$.

However, if $f: X \to Y$ is injective, then the direct image operator $S \mapsto f(S)$ does preserve intersections of subsets $(S_i)_{i \in I}$ indexed over nonempty sets $I$. This is not difficult to check once we verify that

• For any map $f: X \to Y$ (not necessarily injective) and any subset $T \subseteq Y$, we have $f(X) \cap T = f(f^{-1}(T))$.

• If $f$ is injective, then for any subset $S \subseteq X$ we have $S = f^{-1}(f(S))$.

For then, if $T = \bigcap_{i \in I} f(S_i)$ and $I$ is inhabited, we have for some $i$ that $T \subseteq f(S_i) \subseteq f(X)$, and then

$$\array{ T = f(X) \cap T & = & f(f^{-1}(T)) \\ & = & f(f^{-1}(\bigcap_i f(S_i))) \\ & = & f\left(\bigcap_{i \in I} f^{-1}(f(S_i))\right) \\ & = & f\left(\bigcap_{i \in I} S_i\right) }$$

which was to be shown.

Inverse images preserve unions, codirect images preserve intersections

As for the statement that inverse images of unions are unions of inverse images, it turns out this follows from a second fundamental logical adjunction. For a function $f: X \to Y$ and a subset $U \subseteq X$, define

$$\forall_f(U) = \{y \in Y: (\forall_{x: f(x) = y})\; x \in U\}$$

which is analogous to the formula $\exists_f(S) = f(S) = \{y \in Y: (\exists_{x: f(x) = y})\; x \in S\}$ in Remark . (Alternatively, in classical logic where the negation operator $\neg$ on power sets means complementation, we have $\forall_f = \neg \exists_f \neg$.) Then it is easy to verify a second adjunction

$$f^{-1}(T) \subseteq U \qquad iff \qquad T \subseteq \forall_f (U)$$

The same adjointness proof as used to prove property 1., then shows that the left adjoint part which here is the inverse image operator $f^{-1} = f^\ast: P(Y) \to P(X)$ preserves unions, and that right adjoint part which here is the “codirect image” operator $\forall_f: P(X) \to P(Y)$ preserves intersections.

Remark

In terms of some more category theory jargon the above situation may be phrased as follows:

Subsets of some set $S$ are the (-1)-truncated objects in the slice category $Set_{/S}$ of Set over $S$. Since Set is a topos, every morphism, i.e. function $X \longrightarrow Y$ induces a base change adjoint triple

$$Set_{/X} \underoverset {\overset{f_\ast}{\longrightarrow}} {\overset{f_!}{\longrightarrow}} {\overset{f^\ast}{\longleftarrow}} Set_{/Y} \,.$$

Here in the language of type theory

1. the left adjoint $f_!$ is also called the dependent sum

2. the middle morphism is also called context extension

3. the right adjoint $f_\ast$ is also called the dependent product.

All this restricts to the (-1)-truncated objects by composing the left adjoint with (-1)-truncation. Then one also says that we are looking at a morphism in a hyperdoctrine and (under “propositions as types”) may think of

1. the left adjoint $f_!$ as the existential quantifier $\exists_f$

2. the right adjoint $f_\ast$ as the universal quantifier $\forall_f$.

Now the fact that left adjoint $\exists_f$ preserves unions is the fact that left adjoints preserve colimits, and the fact that the left- and right adjoint $f^\ast$ preserves unions and intersections is that in addition right adjoints preserve limits.

Projection formula

Another useful formula for interactions between images and pre-images and intersections is the projection formula:

Proposition

Let $f \,\colon\, X \to Y$ be a function, and let $S \subseteq X,\; T \subseteq Y$ be subsets. Then

$$f\big( S \cap f^{-1}(T) \big) \;=\; f(S) \cap T \,.$$

(e.g. Lee 2000, Ex. A.4(k))

Proof

First observe that we have an inclusion$f(S \cap f^{-1}(T)) \subseteq f(S) \cap T$:

By the defining property of intersections, it suffices to show (i) $f\big(S \cap f^{-1}(T)\big) \subseteq f(S)$ and (ii) $f\big(S \cap f^{-1}(T)\big) \subset T$. Here (i) follows since $S \cap f^{-1}(T) \subseteq S$, and $A \subseteq B$ implies $f(A) \subseteq f(B)$. and (ii) is seen as

$$f\big( S \cap f^{-1}(T) \big) \;\subseteq\; f\big( f^{-1}(T) \big) \;\subseteq\; T \,,$$

where the first step is the functoriality of images with respect to subset inclusions, and the second step is the adjunct of $f^{-1}(T) \subseteq f^{-1}(T)$ with respect to the adjunction between direct and inverse image.

Now to see that this inclusion is actually an equality…

Beck-Chevalley condition

The two ways of consecutively forming images and preimages along maps that form a pullback diagram coincide according to Prop. below – a simple example of what is known as a “Beck-Chevalley condition”.

First to recall our notation: As above, for a set $S \,\in\, Set$, we write $\mathcal{P}(S)$ for its power set regarded as a poset, meaning that

1. $\mathcal{P}(S)$ is the set of subsets $A \subset S$,

2. regarded as a category by declaring that there is a morphisms $A \to B$ precisely if $A \subset B$ as subsets of $S$.

Then for $f \,\colon\, S \longrightarrow T$ a function between sets, we obtain the follows three functors between their power sets:

1. $f_! \,\coloneqq\, f(-) \,\colon\, \mathcal{P}(S) \longrightarrow \mathcal{P}(T)$

   forms images under $f$

2. $f^\ast \,\coloneqq\, f^{-1} \,\colon\, \mathcal{P}(T) \longrightarrow \mathcal{P}(S)$

   forms preimages under $f$,

3. $f_\ast \,\coloneqq\, T \setminus f\big(S \setminus (-)\big) \,\colon\, \mathcal{P}(S) \longrightarrow \mathcal{P}(T)$

   forms the complement of images of complements.

As discussed above, these three functors constitute an adjoint triple

$$f(-) \;\dashv\; f^{-1} \;\dashv\; f_\ast$$

In particular, the familiar/evident fact that forming images and preimages preserves unions may be understood as a special case of the fact that left adjoints preserve colimits.

Proposition

(Beck-Chevalley condition for pre-/images)
Given a pullback diagram in Set

(1)$$\array{ D' &\overset{\beta}{\longrightarrow}& D \\ \mathllap{{}^{\psi}}\big\downarrow &{}^{{}_{(pb)}}& \big\downarrow\mathrlap{{}^{\phi}} \\ C' &\underset{\alpha}{\longrightarrow}& C }$$

then consecutively forming (pre)images along these maps satisfies a Beck-Chevalley condition in that the following diagram of poset-homomorphisms (functors) commutes:

(2)$$\array{ \mathcal{P}(D') &\overset{\beta_!}{\longrightarrow}& \mathcal{P}(D) \\ \mathllap{{}^{\psi^{-1}}}\big\uparrow && \big\uparrow\mathrlap{{}^{\phi^{-1}}} \\ \mathcal{P}(C') &\underset{\alpha_!}{\longrightarrow}& \mathcal{P}(C) }$$

Proof

To see this it is sufficient to check commutativity on singleton-subsets (because every other subset is a union of singletons and, as just remarked, images and preimages preserve unions) and given a singleton set $\{c'\} \subset C'$ on an element $c' \,\in\, C'$, we directly compute as follows:

(3)$$\begin{array}{l} \beta \circ \psi^{-1}\big(\{c'\}\big) \\ \;\simeq\; \beta\Big( \big\{ (c',d) \,\big\vert\, d \,\in\, D \,, \phi(d) = \alpha(c') \big\} \Big) \\ \;=\; \big\{ d \,\in\, D \,\big\vert\, \phi(d) = \alpha(c') \big\} \\ \;=\; \phi^{-1}\big( \{\alpha(c')\} \big) \\ \;=\; \phi^{-1} \circ \alpha\big(\{c'\}\big) \,. \end{array}$$

Here it is the first two steps which make use of the assumption that (1) is a pullback square, namely which means that $D'$ is compatibly bijective to the fiber product of $C'$ with $D$ over $C$:

$$\array{ D' &\simeq& \big\{ (c', d) \in C' \times D \,\big\vert\, \phi(d) \,=\, \alpha(c') \big\} \\ \mathllap{{}^{\beta}} \big\downarrow && \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \big\downarrow\mathrlap{{}^{ (c',d) \mapsto d }} \\ D &=& \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! D }$$

Related statements

• countable unions of countable sets are countable

• limits and colimits by example

References

Textbook accounts include:

• Tom Apostol, p. 44 in: Mathematical Analysis 1973 (pdf)

• John M. Lee, Exercise A.4 in: Introduction to topological manifolds. Graduate Texts in Mathematics 202 (2000), Springer. ISBN: 0-387-98759-2, 0-387-95026-5.

  Second edition: Springer, 2011. ISBN: 978-1-4419-7939-1 (doi:10.1007/978-1-4419-7940-7, errata pdf)

The slogan of Lawvere on logic and adjoint functors appears in

• F. William Lawvere, An elementary theory of the category of sets, Proceedings of the National Academy of Science of the U.S.A 52 (1965), 1506-1511. Reprinted in an expanded version, with commentary by Colin McLarty and by Lawvere, in Reprints in Theory and Applications of Categories, No. 11 (2005), pp. 1-35. (link)

based on

• William Lawvere, Adjointness in Foundations, (TAC), Dialectica 23 (1969), 281-296

The story of Lebesgue’s slip that projections commute with intersections, in attempting to prove that projections of Borel sets are Borel, is well explained here:

• Dave L. Renfro, Mikhail Y. Suslin and Lebesgue’s error, two Mathforum posts dated July 29, 2006 (first post here).

The analogy between adjoint functors and adjoints in linear algebra becomes very strong when jacking up from Hilbert spaces to 2-Hilbert spaces. See

• John C. Baez, Higher-Dimensional Algebra II. 2-Hilbert Spaces, Advances in Mathematics, Volume 127 Issue 2 (10 May 1997), 125-189. https://doi.org/10.1006/aima.1997.1617 (Elsevier link)

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1. There is a famous story of how Suslin discovered an error in an argument of Lebesgue, concerning just this type of routine set-theoretic detail (that Lebesgue perhaps just misremembered). In response, Suslin was led to develop some of the basics of descriptive set theory. Details of the story have been told by Dave Renfro, here and here ↩