nyuuzyou/wikis huggingface dataset:
Predecessor is Infimum

== Theorem ==
Let $\struct{S, \preceq}$ be an ordered set.
Let $x, y \in S$.
Then:
:$x \preceq y$ $x = \map \inf {x, y}$
where $\map \inf {x, y}$ is the infimum of $x$ and $y$
== Proof ==
This is the dual statement of Successor is Supremum by Dual Pairs (Order Theory).
The result follows from the Duality Principle.
Category:Orderings
Category:Infima
Predicate/Examples/Favourite Music

== Examples of Predicates ==
The string Prime.mover is a constant.
The statement:
:$x$ is one of Prime.mover's favourite pieces of music
is a predicate which involves a variable and a constant.
== Sources ==
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1.1$: 'You talking to me?': Example $1.1.3$
Category:Examples of Predicates
Prefix/Examples/Arbitrary Example 1

== Examples of Prefixes ==
Let $abc$ be a string.
Then the prefixes of $abc$ are:
:$\epsilon$, $a$, $ab$ and $abc$
where $\epsilon$ is the null string.
== Sources ==
* : Chapter $1$: Preliminaries: $1.1$ Strings, Alphabets and Languages
Category:Examples of Prefixes
Prefix of Fibonacci String

== Theorem ==
Let $n \in \Z_{>1}$.
Let $S_n$ denote the $n$th Fibonacci string.
Let $m \in \Z$ such that $1 < m \le n$.
Let $F_m$ denote the $m$th Fibonacci number.
The prefix of $S_n$ of length $F_m$ is the Fibonacci string $S_m$.
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:for all $m \in \Z$ such that $1 < m \le n$, the prefix of $S_n$ of length $F_m$ is $S_m$.
=== Basis for the Induction ===
$\map P 3$ is the case:
:$S_3 = \text{ba}$
For $m = 2$, $S_3$ starts with $S_2 = \text{b}$, which is the prefix of $S_3$ of length $F_2 = 1$.
Thus $\map P 3$ is seen to hold.
$\map P 4$ is the case:
:$S_4 = \text{bab}$
For $m = 2$, $S_4$ starts with $S_2 = \text{b}$, which is the prefix of $S_4$ of length $F_2 = 1$.
For $m = 3$, $S_4$ starts with $S_3 = \text{ba}$, which is the prefix of $S_4$ of length $F_3 = 2$.
Thus $\map P 4$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $4 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:for all $m \in \Z$ such that $1 < m \le k$, the prefix of $S_k$ of length $F_m$ is $S_m$.
from which it is to be shown that:
:for all $m \in \Z$ such that $1 < m \le k + 1$, the prefix of $S_{k + 1}$ of length $F_m$ is $S_m$.
=== Induction Step ===
This is the induction step:
By definition of Fibonacci string:
:$S_{k + 1} = S_k S_{k - 1}$
concatenated.
Thus the prefix of $S_{k + 1}$ of length $F_k$ is $S_k$.
By the induction hypothesis, for all $m \in \Z$ such that $1 < m < k - 1$, the prefix of $S_k$ of length $F_m$ is $S_m$.
But we also have that the prefix of $S_{k + 1}$ of length $F_k$ is $S_k$. 
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:for all $m \in \Z$ such that $1 < m \le n$, the prefix of $S_n$ of length $F_m$ is $S_m$.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $36$
Category:Fibonacci Strings
Prefix of String is Substring

== Theorem ==
Let $S$ be a string.
Let $T$ be a prefix of $S$.
Then $T$ is a substring of $S$.
== Proof ==
By definition of substring, there exists a string $T'$ such that:
:$S = TT'$
Hence $S$ is the concatenation of the null string, $T$, and $T'$.
Thus by definition of substring, $T$ is a substring of $S$.
Category:Substrings
Prefix of WFF of Predicate Logic is not WFF

== Theorem ==
Let $\mathbf A$ be a WFF of predicate logic.
Let $\mathbf S$ be a prefix of $\mathbf A$.
Then $\mathbf S$ is not a WFF of predicate logic.
== Proof ==
Let $\map l {\mathbf Q}$ denote the length of a string $\mathbf Q$.
By definition, $\mathbf S$ is a prefix of $\mathbf A$ iff $\mathbf A = \mathbf{ST}$ for some non-null string $\mathbf T$.
Thus we note that $\map l {\mathbf S} < \map l {\mathbf A}$.
The proof proceeds by induction on $\map l {\mathbf A}$.
=== Basis for the Induction ===
Let $\mathbf A$ be a WFF such that $\map l {\mathbf A} = 1$.
Then for a prefix $\mathbf S$:
:$\map l {\mathbf S} < 1 = 0$
That is, $\mathbf S$ must be the null string, which is not a WFF.
So the result holds for WFFs of length $1$.
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$ with $n \ge n_0$.
Assume the result holds for all WFFs of length $k$ or less.
This is our induction hypothesis.
=== Induction Step ===
This is our induction step:
Let $\mathbf A$ be a WFF such that $\map l {\mathbf A} = k + 1$.
Suppose $\mathbf D$ is a prefix of $\mathbf A$ which happens to be a WFF.
That is, $\mathbf A = \mathbf{DT}$ where $\mathbf T$ is non-null.
We need to investigate the following cases:
:$(1): \quad \mathbf A = \neg \mathbf B$, where $\mathbf B$ is a WFF of length $k$.
:$(2): \quad \mathbf A = \left({\mathbf B \circ \mathbf C}\right)$ where $\circ$ is one of the binary connectives.
:$(3): \quad \mathbf A = \map p {t_1, t_2, \ldots, t_n}$, where $t_1, t_2, \ldots, t_n$ are terms, and $p \in \PP_n$.
:$(4): \quad \mathbf A = ( Q x: \mathbf B )$, where $\mathbf B$ is a WFF of length $k-5$, $Q$ is a quantifier ($\forall$ or $\exists$) and $x$ is a variable.
We deal with these one by one.
Cases $(1)$ and $(2)$ are covered by the argument in Prefix of WFF of PropLog is not WFF.
:$(3): \quad$ The atomic WFF $\mathbf A = \map p {u_1, u_2, \ldots, u_n}$:
Here we have that $\mathbf D$ would be a string of the form:
* $p$ where $p$ is an $n$-ary predicate symbol. This can not be a WFF.
* $p ($ which is also not a WFF.
* $p (u_1, u_2, \ldots, u_k$ which can not be a WFF.
* $p (u_1, u_2, \ldots, u_k,$ which can not be a WFF.
:$(4): \quad \mathbf A = ( Q x: \mathbf B )$:
$\mathbf D$ can not be $($, $( Q$, $( Q x$ or $( Q x:$ as none of these are WFFs.
Thus $\mathbf D$ is a WFF, starting with $( Q x: $; hence $\mathbf D = ( Q x: \mathbf E )$, where $\mathbf E$ is also a WFF.
We remove the initial $( Q x: $ and the last closing parenthesis $)$ from $\mathbf A = \mathbf{DT}$ to get $\mathbf B = \mathbf{ET}'$ (where $\mathbf T'$ results from $\mathbf T$ by removing the last closing parenthesis).
Recall that $\mathbf B$ is a WFF of length $k - 5$.
It has $\mathbf E$ as a prefix, which is itself a WFF.
This contradicts the induction hypothesis.
Therefore no prefix of $\mathbf A = ( Q x: \mathbf{B} )$ can be a WFF.
Thus all four cases have been investigated, and we have found that no prefix of any WFF of length $k + 1$ can be a WFF.
The result follows by the Second Principle of Mathematical Induction.
== Sources ==
* : $\S 2.2$: Lemma $2.2.3$
Category:Predicate Logic
Preimage Theorem

== Theorem ==
Let $y$ be a regular value of a smooth submersion $f: X \to Y$.
Then the preimage $\map {f^{-1} } y$ is a smooth submanifold of $X$, with $\dim \map {f^{-1} } y = \dim X - \dim Y$.
== Proof ==
Let $k, l$ be natural numbers with $k \ge l$.
By the Local Submersion Theorem, there exists coordinates in some open neighborhoods of $x, y$ such that $\map f {x_1, x_2, \ldots, x_k} = \tuple {x_1, \ldots, x_l}$ and $y$ corresponds to $\tuple {0, \ldots, 0}$.
Let $V$ be that neighborhood of $x$.
Then $\map {f^{-1} } y \cap V$ is the set of points where $x_1 = 0, \ldots, x_l = 0$.
The functions $x_{l + 1}, \ldots, x_k$ therefore form a coordinate system on the set $\map {f^{-1} } y \cap V$, which is a relatively open subset of $\map {f^{-1} } y$.
Together these functions then form a diffeomorphism to a Euclidean space.
We also have, by the regular value properties of $y$, a surjection of tangent spaces from $x$ to $y$.
This ensures smoothness of the solution set $\map {f^{-1} } y$.
== Also known as ==
This theorem is also known as the submersion level set theorem, regular value theorem and regular level set theorem.
== Sources ==
* : $5$: Submanifolds $\S$ Embedded Submanifolds
Category:Named Theorems
Category:Manifolds
Preimage is Subset Iff Subset of Complement of Image of Complement

== Theorem ==
Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall A \subseteq S, B \subseteq T : f^{-1} \sqbrk B \subseteq A$ $B \subseteq T \setminus f \sqbrk {S \setminus A}$
== Proof ==
We have:
Category:Images
Category:Preimages
Preimage of Composite Relation

== Theorem ==
Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be relations.
Let $\RR_2 \circ \RR_1 \subseteq S_1 \times T_2$ be the composition of $\RR_1$ and $\RR_2$.
Then the preimage of $\RR_2 \circ \RR_1$ is given by:
:$\Preimg {\RR_2 \circ \RR_1} = \Preimg {\Img {\RR_1} \cap \Preimg {\RR_2} }$
== Proof ==
Category:Composite Relations
Preimage of Cover is Cover

== Theorem ==
Let $\phi: S \to T$ be a mapping between the sets $S$ and $T$.
Let $\UU$ be a cover of $T$.
Then the set:
:$\set {\map {\phi^{-1} } U: U \in \UU}$
is a cover of $S$.
== Proof ==
Let $x \in S$.
Then $\map \phi x \in T$.
Since $\UU$ is a cover of $T$:
:$\exists U \in \UU: \map \phi x \in U$
By definition of preimage, $x \in \map {\phi^{-1} } U$.
So:
:$\forall x \in S: \exists \map {\phi^{-1} } U \in S: x \in \map {\phi^{-1} } U$
That is, $\set {\map {\phi^{-1} } U: U \in \UU}$ is a cover of $S$.
Category:Preimages under Mappings
Category:Covers
Preimage of Dilation of Set under Linear Transformation is Dilation of Preimage

== Theorem ==
Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $\lambda \in K$. 
Then: 
:$T^{-1} \sqbrk {\lambda E} = \lambda T^{-1} \sqbrk E$
where $\lambda E$ denotes the dilation of $E$ by $\lambda$.
== Proof ==
The result is immediate when $\lambda = 0_K$, since $T^{-1} \sqbrk {\set { {\mathbf 0}_X} } = \set { {\mathbf 0}_X}$.
Now take $\lambda \ne 0_K$.
Let $x \in X$.
We have:
:$x \in T^{-1} \sqbrk {\lambda E}$
:
:$T x \in \lambda E$
From linearity, this is the case :
:$\map T {\lambda^{-1} x} \in E$
This is equivalent to:
:$\lambda^{-1} x \in T^{-1} \sqbrk E$
Which in turn is equivalent to:
:$x \in \lambda T^{-1} \sqbrk E$
So we obtain:
:$T^{-1} \sqbrk {\lambda E} = \lambda T^{-1} \sqbrk E$
Category:Dilation of Subsets of Vector Spaces
Preimage of Element under Projection

== Theorem ==
Let $A$ and $B$ be sets.
Let $A \times B$ be the cartesian product of $A$ and $B$.
Let $\pr_1: A \times B \to A$ be the first projection of $A \times B$.
Let $a \in A$.
Then:
:$\pr_1^{-1} \sqbrk {\set a} = \set {\tuple {a, b}: b \in B}$
that is:
:$\pr_1^{-1} \sqbrk {\set a} = \set a \times B$
== Proof ==
Directly apparent from the definition of cartesian product.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $5$
Category:Projections
Preimage of Horizontal Section of Function is Horizontal Section of Preimage

== Definition ==
Let $X$ and $Y$ be sets.
Let $f : X \times Y \to \overline \R$ be an extended real-valued function. 
Let $y \in Y$.
Let $D \subseteq \R$. 
Then: 
:$\paren {f^y}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}^y$
where:
:$f^y$ is the $y$-horizontal section of $f$
:$\paren {f^{-1} \sqbrk D}^y$ is the $y$-horizontal section of $f^{-1} \sqbrk D$.
== Proof ==
Note that:
:$x \in \paren {f^y}^{-1} \sqbrk D$
: 
:$\map {f^y} x \in D$
from the definition of preimage.
That is, by the definition of the $y$-horizontal section: 
:$\map f {x, y} \in D$
From the definition of preimage, this is equivalent to: 
:$\tuple {x, y} \in f^{-1} \sqbrk D$
Which in turn is equivalent to: 
:$x \in \paren {f^{-1} \sqbrk D}^y$
from the definition of the $y$-horizontal section.
So:
:$x \in \paren {f^y}^{-1} \sqbrk D$ $x \in \paren {f^{-1} \sqbrk D}^y$.
giving:
:$\paren {f^y}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}^y$
Category:Horizontal Section of Functions
Category:Horizontal Section of Sets
Category:Preimages under Mappings
Preimage of Ideal under Ring Homomorphism is Ideal

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S_2$ be an ideal of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is an ideal of $R_1$ such that $\map \ker \phi \subseteq S_1$.
== Proof ==
From Preimage of Subring under Ring Homomorphism is Subring‎ we have that $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$.
We now need to show that $S_1$ is an ideal of $R_1$.
Let $s_1 \in S_1, r_1 \in R_1$.
Then:
Thus:
:$r_1 \circ_1 s_1 \in \phi^{-1} \sqbrk {S_2} = S_1$
Similarly for $s_1 \circ_1 r_1$.
So $S_1$ is an ideal of $R_1$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 3^\circ$
Category:Ring Homomorphisms
Category:Ideal Theory
Preimage of Image of Ideal under Ring Homomorphism

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $K = \map \ker \phi$, be the kernel of $\phi$. 
Let $J$ be an ideal of $R_1$.
Then:
:$\phi^{-1} \sqbrk {\phi \sqbrk J} = J + K$
== Proof ==
As an ideal is a subring, the result Preimage of Image of Subring under Ring Homomorphism applies directly.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 2^\circ$
Category:Ring Homomorphisms
Category:Ideal Theory
Preimage of Image of Linear Transformation

== Theorem ==
Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $A \subseteq X$ be a non-empty set.
Then $T^{-1} \sqbrk {T \sqbrk A} = \ker T + A$
== Proof ==
Let $x \in T^{-1} \sqbrk {T \sqbrk A}$.
Then $T x \in T \sqbrk A$.
Then there exists $y \in A$ such that $T x = T y$. 
Hence we have $\map T {x - y} = 0$.
Hence $x - y \in \ker T$.
So $x \in y + \ker T$.
We have $y + \ker T \subseteq A + \ker T$, so we obtain:
:if $x \in \in T^{-1} \sqbrk {T \sqbrk A}$ then $x \in \ker T + A$
so that:
:$T^{-1} \sqbrk {T \sqbrk A} \subseteq \ker T + A$
Now let $x \in \ker T + A$.
Then there exists $v \in \ker T$ and $y \in A$ such that $x = v + y$.
From linearity, we have $T x = \map T {v + y} = T v + T y$.
Since $v \in \ker T$, we have $T x = T y$.
Since $T y \in T \sqbrk A$, we have $T x \in T \sqbrk A$.
So $x \in T^{-1} \sqbrk {T \sqbrk A}$.
Category:Linear Transformations
Preimage of Image of Subgroup under Group Epimorphism

== Theorem ==
Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group epimorphism.
Let $K = \map \ker \phi$ denote the kernel of $\phi$.
Then:
:$\phi^{-1} \sqbrk {\phi \sqbrk H} = H \circ K$
where:
:$\phi \sqbrk H$ denotes the image of $H$ under $\phi$
:$\phi^{-1} \sqbrk {\phi \sqbrk H}$ denotes the preimage of $\phi \sqbrk H$ under $\phi$
:$H \circ K$ denotes the subset product of $H$ with $K$.
== Proof ==
Let $e_2$ be the identity element of $\struct {G_2, *}$.
Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk H}$.
Then:
So we have shown that:
:$\phi^{-1} \sqbrk {\phi \sqbrk H} \subseteq H \circ K$
Now suppose that $x \in H \circ K$.
Then:
So we have shown that:
:$H \circ K \subseteq \phi^{-1} \sqbrk {\phi \sqbrk H}$
Hence the result by definition of set equality.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(a)}$
Category:Group Epimorphisms
Category:Subgroups
Preimage of Image of Subring under Ring Homomorphism

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $K = \map \ker \phi$ be the kernel of $\phi$. 
Let $J$ be a subring of $R_1$.
Then:
:$\phi^{-1} \sqbrk {\phi \sqbrk J} = J + K$
== Proof ==
Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk J}$.
Then:
So we have shown that:
:$\phi^{-1} \sqbrk {\phi \sqbrk J} \subseteq J + K$
Now suppose that $x \in J + K$.
Then:
So we have shown that:
:$J + K \subseteq \phi^{-1} \sqbrk {\phi \sqbrk J}$
Hence the result by definition of set equality.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 2^\circ$
Category:Ring Homomorphisms
Category:Subrings
Preimage of Image of Subset under Injection equals Subset

== Theorem ==
Let $f: S \to T$ be an injection.
Then:
:$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$
where:
:$f \sqbrk A$ denotes the image of $A$ under $f$
:$f^{-1}$ denotes the inverse of $f$
:$f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.
== Proof ==
Let $f$ be an injection.
From Subset of Domain is Subset of Preimage of Image, we have that:
:$\forall A \subseteq S: A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
by dint of $f$ being a relation.
So what we need to do is show that:
:$\forall A \subseteq S: \paren {f^{-1} \circ f} \sqbrk A \subseteq A$
Take any $A \subseteq S$.
Let $x \in A$.
We have:
Thus we see that:
:$\paren {f^{-1} \circ f} \sqbrk A \subseteq A$
and hence the result:
:$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$
== Also see ==
* Subset equals Preimage of Image implies Injection
* Subset equals Preimage of Image iff Mapping is Injection
== Sources ==
* : $\S 10$: Inverses and Composites
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $1 \ \text{(c})$
* : $1$: Set Theory and Logic: $\S 2$: Functions
Category:Injections
Category:Composite Mappings
Category:Preimages under Mappings
Preimage of Image under Left-Total Relation is Superset

== Theorem ==
Let $\RR \subseteq S \times T$ be a left-total relation.
Then:
:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$
where:
:$\RR \sqbrk A$ denotes the image of $A$ under $\RR$
:$\RR^{-1} \sqbrk A$ denotes the preimage of $A$ under $\RR$
:$\RR^{-1} \circ \RR$ denotes composition of $\RR^{-1}$ and $\RR$.
This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
:$\forall A \in \powerset S: A \subseteq \map {\paren {\RR^\gets \circ \RR^\to} } A$
== Proof ==
Suppose $A \subseteq S$.
We have:
Then we have that $\RR$ is a left-total relation.
Thus:
:$\exists t \in T: \tuple {x, t} \in \RR$
Hence by definition of inverse relation:
:$\exists t \in T: \tuple {t, x} \in \RR^{-1}$
Hence:
So by definition of subset:
:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$
== Also see ==
* Subset of Domain is Subset of Preimage of Image
Category:Preimages under Relations
Category:Composite Relations
Category:Left-Total Relations
Preimage of Intersection under Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cap T_2} = \map {f^\gets} {T_1} \cap \map {f^\gets} {T_2}$
=== General Result ===
=== Family of Sets ===
== Proof ==
As $f$, being a mapping, is also a many-to-one relation, it follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.
Thus Image of Intersection under One-to-Many Relation applies:
:$\RR \sqbrk {T_1 \cap T_2} = \RR \sqbrk {T_1} \cap \RR \sqbrk {T_2}$
where here $\RR = f^{-1}$.
== Also see ==
* Image of Intersection under Mapping
* Preimage of Intersection under Relation
* Image of Union under Mapping
* Preimage of Union under Mapping
== Sources ==
* : Chapter $0$: Functions: Theorem $7 \ \text{(c)}$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(b)}$
* : $\S 1.3$: Functions and mappings. Images and preimages: Theorem $2$
* : Chapter $1$: Mappings: $\S 12 \alpha$
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.4$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(jjj)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(c)}$
* : $\S 2$: Problem $4 \ \text{(ii)}$
Category:Preimages under Mappings
Category:Set Intersection
Category:Preimage of Intersection under Mapping
Preimage of Intersection under Mapping/Family of Sets

== Theorem ==
Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
where:
:$\ds \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\family {T_i}_{i \mathop \in I}$.
:$f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.3$: Functions and mappings. Images and preimages: Remark $2$
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.6 \ \text{(c)}$
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $1$
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $4 \ \text{(b})$
* : Notation and Terminology
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.3 \ \text{(ii)}$
* : Appendix $\text{A}.3$: Functions: Problem $\text{A}.3.1$
* : $\S 2$
Category:Indexed Families
Category:Preimage of Intersection under Mapping
Preimage of Intersection under Mapping/Family of Sets/Proof 1

== Theorem ==
== Proof ==
As $f$ is a mapping, it is by definition also a many-to-one relation.
It follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.
Thus Image of Intersection under One-to-Many Relation: Family of Sets can be applied for $\RR = f^{-1}$:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {T_i}$
where $\RR \sqbrk {T_i}$ denotes the image of $T_i$ under $\RR$.
Category:Preimage of Intersection under Mapping
Preimage of Intersection under Mapping/General Result

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds f^{-1} \sqbrk {\bigcap \mathbb T} = \bigcap_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$ 
== Proof ==
$f$ is a mapping.
Therefore it is by definition a many-to-one relation.
It follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.
Thus Image of Intersection under One-to-Many Relation: General Result applies:
:$\ds \RR \sqbrk {\bigcap \mathbb T} = \bigcap_{X \mathop \in \mathbb T} \RR \sqbrk X$
where here $\RR = f^{-1}$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $6 \ \text {(b)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.3$
Category:Preimage of Intersection under Mapping
Preimage of Intersection under Relation

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $C$ and $D$ be subsets of $T$.
Then:
:$\RR^{-1} \sqbrk {C \cap D} \subseteq \RR^{-1} \sqbrk C \cap \RR^{-1} \sqbrk D$
=== General Result ===
=== Family of Sets ===
== Also see ==
* Image of Intersection under Relation
* Image of Union under Relation
* Preimage of Union under Relation
== Proof ==
This follows from Image of Intersection under Relation, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules.
Category:Relation Theory
Category:Set Intersection
Preimage of Intersection under Relation/Family of Sets

== Theorem ==
Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$\ds \RR^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} \subseteq \bigcap_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where $\ds \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\family {T_i}_{i \mathop \in I}$.
== Proof ==
This follows from Image of Intersection under Relation: Family of Sets, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules.
Category:Relation Theory
Category:Set Intersection