nyuuzyou/wikis huggingface dataset:
Preimage of Subset under Inclusion Mapping

== Theorem ==
Let $S$ be a set.
Let $H \subseteq S$ be a subset of $S$.
Let $i_H: H \to S$ be the inclusion mapping on $H$.
Let $T \subseteq S$.
Then:
:$i_H^{-1} \sqbrk T = T \cap H$
where $i_H^{-1} \sqbrk T$ is the preimage of $T$ under $i_H$.
== Proof ==
By definition of preimage:
:$i_H^{-1} \sqbrk T = \set {h \in H: \exists t \in T: \map {i_H} h = t}$
Let $T \cap H = \O$.
Then:
:$\nexists h \in H: \exists t \in T: \map {i_H} h = t$
That is:
:$i_H^{-1} \sqbrk T = \O$
That is:
:$i_H^{-1} \sqbrk T = T \cap H$
Let $T \cap H \ne \O$.
From Intersection is Subset:
:$H \ne \O$
and:
:$T \ne \O$
From Subset of Empty Set:
:$S \ne \O$
Suppose $i_H^{-1} \sqbrk T = \O$.
Then from Empty Set is Subset of All Sets:
:$i_H^{-1} \sqbrk T \subseteq T \cap H$
Otherwise $i_H^{-1} \sqbrk T \ne \O$.
Let $x \in i_H^{-1} \sqbrk T$.
By the definition of inclusion mapping:
:$\map {i_H} x = x$
By definition of preimage of $T$ under $i_H$:
:$\map {i_H} x \in T$
Thus $x \in T$.
By definition of inclusion mapping:
:$x \in \Dom {i_H} = H$
So $x \in T$ and $x \in H$ and so by definition of set intersection:
:$x \in T \cap H$
Thus by definition of subset:
:$i_H^{-1} \sqbrk T \subseteq T \cap H$
Now let $x \in T \cap H$.
As $H = \Dom {i_H}$ it follows that:
:$x \in \Dom {i_H}$
As $x \in T \cap H$ it follows by definition of set intersection that $x \in T$.
Thus by definition of inclusion mapping again:
:$\map {i_H} x \in T$
and so by definition of preimage of $T$:
:$x \in i_H^{-1} \sqbrk T$
Therefore:
:$x \in T \cap H \implies x \in i_H^{-1} \sqbrk T$
Hence, by definition of subset:
:$T \cap H \subseteq i_H^{-1} \sqbrk T$
We have shown that:
:$i_H^{-1} \sqbrk T \subseteq T \cap H$
and:
:$T \cap H \subseteq i_H^{-1} \sqbrk T$
Hence by definition of set equality:
:$T \subseteq S: i_H^{-1} \sqbrk T = T \cap H$
Category:Inclusion Mappings
Category:Subsets
Preimage of Subset under Mapping/Examples

== Examples of Preimage of Subset under Mapping ==
=== Subset of Image of Square Root Function ===
=== Identity Function with Discontinuity ===
=== Preimage of $\closedint {-2} 0$ under $\map f x = x^2 - x - 2$ ===
=== Preimage of $\set 0$ under $\map f x = x^2 - x - 2$ ===
=== Preimage of $\closedint {-5} {-4}$ under $\map f x = x^2 - x - 2$ ===
=== Preimages of $\map f {x, y} = x y$ ===
=== Preimages of $\map g {x, y} = \tuple {x^2 + y^2, x y}$ ===
Category:Examples of Preimages
Preimage of Subset under Mapping/Examples/Identity Function with Discontinuity

== Example of Preimage of Subset under Mapping ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$
Let the following subsets of $\R$ be defined:
Then the preimages under $f$ of these sets is:
:$f^{-1} \sqbrk A = \set {x \in \R: \map f x < 0} = \openint \gets 0$
:$f^{-1} \sqbrk B = \set {x \in \R: \map f x > 0} = \openint 0 \to$
:$f^{-1} \sqbrk C = \set {x \in \R: 0 < \map f x < 1} = \openint 0 1$
:$f^{-1} \sqbrk D = \set {x \in \R: -1 < \map f x < 1} = \openint {-1} 0 \cup \openint 0 1$
:$f^{-1} \sqbrk {E \cup F} = \openint {-2} {-1} \cup \openint {\dfrac 1 2} 2 \cup \set 0$
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.15$
Category:Examples of Preimages
Preimage of Subset under Mapping/Examples/Preimage of -2 to 0 under x^2-x-2

== Example of Preimage of Subset under Mapping ==
Let $f: \R \to \R$ be the mapping defined as:
:$\forall x \in \R: \map f x = x^2 - x - 2$
The preimage of the closed interval $\closedint {-2} 0$ is:
:$f^{-1} \closedint {-2} 0 = \closedint {-1} 0 \cup \closedint 1 2$
== Proof ==
Trivially, by differentiating $x^2 - x - 2$ $x$:
:$f' = 2 x - 1$
and equating $f'$ to $0$, the minimum of $\Img f$ is seen to occur at $\map f {\dfrac 1 2} = -\dfrac 9 4$.
We see that $\closedint {-2} 0$ is well within the codomain of $f$.
Hence we should be able to solve for $\map f x = -2$ and $\map f x = 0$ and get two values for each.
So:
Then:
The result follows.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions
Category:Examples of Preimages
Preimage of Subset under Mapping/Examples/Preimage of -5 to -4 under x^2-x-2

== Example of Preimage of Subset under Mapping ==
Let $f: \R \to \R$ be the mapping defined as:
:$\forall x \in \R: \map f x = x^2 - x - 2$
The preimage of the closed interval $\closedint {-5} {-4}$ is:
:$f^{-1} \closedint {-2} 0 = \O$
the Empty Set
== Proof ==
Trivially, by differentiating $x^2 - x - 2$ $x$:
:$f' = 2 x - 1$
and equating $f'$ to $0$, the minimum of $\Img f$ is seen to occur at $\map f {\dfrac 1 2} = -\dfrac 9 4$.
We see that $\closedint {-5} {-4}$ is well outside the image of $f$.
Hence the result.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions
Category:Examples of Preimages
Preimage of Subset under Mapping/Examples/Preimage of 0 under x^2-x-2

== Example of Preimage of Subset under Mapping ==
Let $f: \R \to \R$ be the mapping defined as:
:$\forall x \in \R: \map f x = x^2 - x - 2$
The preimage of the singleton $\set 0$ is:
:$f^{-1} \sqbrk {\set 0} = \set {-1, 2}$
which is the set of roots of $f$.
== Proof ==
Hence the result.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions
Category:Examples of Preimages
Preimage of Subset under Mapping/Examples/Preimages of f(x, y) = (x^2 + y^2, x y)

== Example of Preimage of Subset under Mapping ==
:
Let $g: \R^2 \to \R^2$ be the mapping defined as:
:$\forall \tuple {x, y} \in \R^2: \map g {x, y} = \tuple {x^2 + y^2, x y}$
Let the following subset of $\R^2$ be defined:
:$S = g^{-1} \sqbrk {\openint 0 3 \times \openint 0 1}$
Then:
:$S = \set {\tuple {x, y} \in \R^2: x y < 1} \cap \set {\tuple {x, y} \in \R^2: 0 < x^2 + y^2 < 3}$
=== Continuity ===
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.16 \ \text {(b)}$
Category:Examples of Preimages
Preimage of Subset under Mapping/Examples/Preimages of f(x, y) = (x^2 + y^2, x y)/Continuity

== Example of Preimage of Subset under Mapping ==
:
Let $g: \R^2 \to \R^2$ be the mapping defined as:
:$\forall \tuple {x, y} \in \R^2: \map g {x, y} = \tuple {x^2 + y^2, x y}$
$g$ is a continuous mapping.
== Proof ==
By Continuous Mapping to Product Space it is sufficient to demonstrate that:
:$h = \pr_1 \circ g$
and:
:$k = \pr_2 \circ g$
where $\pr_1$ and $\pr_2$ are the first projection and seconod projection respectively on $\R^2$, are continuous.
We have:
:$\map k {x, y} = x y$
so by the Product Rule for Continuous Real Functions, $k$ is continuous.
We have:
:$\map h {x, y} = x^2 + y^2$
so by the Product Rule for Continuous Real Functions and the Sum Rule for Continuous Real Functions, $h$ is continuous.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.5$: Products
Category:Examples of Preimages
Preimage of Subset under Mapping/Examples/Preimages of f(x, y) = (x^2 + y^2, x y)/Mistake

== Source Work ==
:
:$2$: Continuity generalized: metric spaces:
::$2.3$: Open sets in metric spaces:
:::Example $2.3.16 \ \text {(b)}$
== Mistake ==
:Let $g: \R^2 \to \R^2$ be given by $\map g {x, y} = \tuple {x^2 + y^2, x y}$. Then for example
:We know from (a) that $\set {\tuple {x, y} \in \R^2: 0 < x y < 1}$ is the shaded region in Fig. 2.2. Also, $\set {\tuple {x, y} \in \R^2: 0 < x^2 + y^2 < 2}$ is the interior of a disc of radius $2$ with its centre removed. Hence $S$, the set of all $\tuple {x, y}$ satisfying both conditions, is the intersection of the shaded region with the outlined disc, not including any of the boundary curves or lines.
:
== Correction ==
The graph as given does not match the definition of the problem.
The graph shows distinct intersections between $x^2 + y^2 = 2$ and $x y = 1$.
But in fact the curves are tangent to each other:
:
In order to make the diagram correct, and to illustrate the point which is being made, it may be suggested that the first curve be $x^2 + y^2 = 3$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.16 \ \text {(b)}$
Category:Mistakes/Introduction to Metric and Topological Spaces
Preimage of Subset under Mapping/Examples/Preimages of f(x, y) = x y

== Example of Preimage of Subset under Mapping ==
:
Let $f: \R^2 \to \R$ be the real function of $2$ variables defined as:
:$\forall \tuple {x, y} \in \R^2: \map f {x, y} = x y$
Let the following subsets of $\R^2$ be defined:
Then:
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.16 \ \text {(a)}$
Category:Examples of Preimages
Preimage of Subset under Mapping/Examples/Subset of Image of Square Root Function

== Example of Preimage of Subset under Mapping ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^2$
Let $A \subseteq \R$ be defined as:
:$A := \closedint 4 9 = \set {x \in \R: 4 \le x \le 9}$
Then the preimage of $A$ under $f$ is:
:$f^{-1} \sqbrk A = \closedint {-3} {-2} \cup \closedint 2 3$
Let $B \subseteq \R$ be defined as:
:$B := \closedint {-9} {-4} = \set {x \in \R: -9 \le x \le -4}$
Then the preimage of $B$ under $f$ is:
:$f^{-1} \sqbrk B = \O$
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Functions
* 
* 
Category:Examples of Preimages
Preimage of Subset under Mapping equals Union of Preimages of Elements

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $f^{-1} \subseteq T \times S$ be the inverse of $f$, defined as:
:$f^{-1} = \set {\tuple {t, s}: \map f s = t}$
Let $Y \subseteq T$ be a subset of $T$.
Then:
:$\ds f^{-1} \sqbrk Y = \bigcup_{y \mathop \in Y} \map {f^{-1} } y$
where:
:$f^{-1} \sqbrk Y$ is the preimage of the subset $Y$ under $f$
:$\map {f^{-1} } y$ is the preimage of the element $y$ under $f$.
== Proof ==
By definition, $f^{-1} \subseteq T \times S$ is a relation on $T \times S$.
Thus Image of Subset under Relation equals Union of Images of Elements can be applied directly.
Category:Preimages under Mappings
Category:Subsets
Category:Set Union
Preimage of Subset under Relation equals Union of Preimages of Elements

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^{-1} \subseteq T \times S$ be the inverse relation to $\RR$
Let $Y \subseteq T$ be a subset of $T$.
Then:
:$\RR^{-1} \sqbrk Y = \ds \bigcup_{y \mathop \in Y} \map {\RR^{-1} } y$
where:
:$\RR^{-1} \sqbrk Y$ is the preimage of the subset $Y$ under $\RR$
:$\RR^{-1} \sqbrk Y$ is the preimage of the element $y$ under $\RR$.
== Proof ==
By definition, $\RR^{-1} \subseteq T \times S$ is a relation on $T \times S$.
Thus Image of Subset under Relation equals Union of Images of Elements can be applied directly.
Category:Relation Theory
Preimage of Union Mapping is Union of Preimages

== Theorem ==
Let $A$, $B$ and $Y$ be sets.
Let $f: A \to Y$ and $g: B \to Y$ be mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$
Let $S \subseteq Y$ be a subset of $Y$.
Then:
:$\paren {f \cup g}^{-1} \sqbrk S = f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$
== Proof ==
From Union of Mappings which Agree is Mapping:
:$f \cup g$ is well-defined
Let $x \in \paren {f \cup g}^{-1} \sqbrk S$
By :
:$\paren {f \cup g}^{-1} \sqbrk S \subseteq A \cup B$
Let $x \in A$.
We have:
:$\map f x = \map {\paren {f \cup g} } x \in S$
By :
:$x \in f^{-1} \sqbrk S \subseteq f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$
Let $x \in B$.
We have:
:$\map g x = \map {\paren {f \cup g} } x \in S$
By :
:$x \in g^{-1} \sqbrk S \subseteq f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$
In both cases: 
:$x \in f^{-1} \sqbrk U \cup g^{-1} \sqbrk S$
By :
:$\paren {f \cup g}^{-1} \sqbrk S \subseteq f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$
Let $x \in f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$.
 consider the case of $x \in f^{-1} \sqbrk S \subseteq A$.
Then:
$\map {\paren {f \cup g} } x = \map f x \in S$
Hence:
:$x \in \paren {f \cup g}^{-1} \sqbrk S$
By :
:$f^{-1} \sqbrk S \cup g^{-1} \sqbrk S \subseteq \paren {f \cup g}^{-1} \sqbrk S$
By :
:$f^{-1} \sqbrk S \cup g^{-1} \sqbrk S = \paren {f \cup g}^{-1} \sqbrk S$
Category:Preimages under Mappings
Category:Union Mappings
Preimage of Union under Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cup T_2} = \map {f^\gets} {T_1} \cup \map {f^\gets} {T_2}$
=== General Result ===
=== Family of Sets ===
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation:
:$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$
== Also see ==
* Image of Union under Mapping
* Image of Intersection under Mapping
* Preimage of Intersection under Mapping
== Sources ==
* : Chapter $0$: Functions: Theorem $7 \ \text{(b)}$
* : 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 $1$
* : Chapter $1$: Mappings: $\S 12 \alpha$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(jj)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(b)}$
Category:Preimage of Union under Mapping
Category:Preimages under Mappings
Category:Set Union
Preimage of Union 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:
:$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
where:
:$\ds \bigcup_{i \mathop \in I} T_i$ denotes the union 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 ==
* : Chapter $0$: Functions: Theorem $7 \ \text{(d)}$
* : $\S 10$: Inverses and Composites
* : $\S 1.3$: 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 $4 \ \text{(a})$
* : Notation and Terminology
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.3 \ \text{(i)}$
* : Appendix $\text{A}.3$: Functions: Problem $\text{A}.3.1$
* : $\S 2$
Category:Preimage of Union under Mapping
Category:Indexed Families
Preimage of Union under Mapping/Family of Sets/Proof 1

== Theorem ==
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: Family of Sets:
:$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where $\RR^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $\RR^{-1}$.
Category:Preimage of Union under Mapping
Preimage of Union under Mapping/Family of Sets/Proof 2

== Theorem ==
== Proof ==
We have that $f$ is a mapping, and so also a relation.
Thus its inverse $f^{-1}$ is also a relation.
Hence we can apply Image of Union under Relation: Family of Sets:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {T_i}$
where $\RR \sqbrk {T_i}$ denotes the image of $T_i$ under $\RR$.
Category:Preimage of Union under Mapping
Preimage of Union 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:
:$\ds f^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: General Result:
:$\ds \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $6 \ \text {(a)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.3$
Category:Preimage of Union under Mapping
Preimage of Union under Relation

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$
=== General Result ===
=== Family of Sets ===
== Proof ==
We have that $\RR^{-1}$ is a relation.
The result follows from Image of Union under Relation.
== Also see ==
* Preimage of Intersection under Relation
* Image of Intersection under Relation
* Image of Union under Relation
Category:Preimages under Relations
Category:Set Union
Category:Preimage of Union under Relation
Preimage of Union 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:
:$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where:
:$\ds \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \mathop \in I}$
:$\RR^{-1} \left[{T_i}\right]$ denotes the preimage of $T_i$ under $\RR$.
== Proof ==
We have that $\RR^{-1}$ is itself a relation.
The result follows from Image of Union under Relation: Family of Sets.
Category:Preimage of Union under Relation
Category:Indexed Families
Preimage of Union under Relation/General Result

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$
where $\RR^{-1} \sqbrk X$ denotes the preimage of $X$ under $\RR$.
== Proof ==
We have that $\RR^{-1}$ is a relation.
The result follows from Image of Union under Relation: General Result.
Category:Preimage of Union under Relation
Preimage of Upper Section under Increasing Mapping is Upper

== Theorem ==
Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be preordered sets.
Let $f: S \to T$ be an increasing mapping.
Let $X \subseteq T$ be a upper subset of $T$.
Then $f^{-1} \sqbrk X$ is upper
where $f^{-1} \sqbrk X$ denotes the preimage of $X$ under $f$.
== Proof ==
Let $x \in f^{-1} \sqbrk X$, $y \in S$ such that
:$x \preceq y$
By definition of increasing mapping:
:$\map f x \precsim \map f y$
By definition of preimage of set:
:$\map f x \in X$
By definition of upper section:
:$\map f y \in X$
Thus by definition of preimage of subset:
:$y \in f^{-1} \sqbrk X$
== Sources ==
* 
Category:Upper Sections
Preimage of Vertical Section of Function is Vertical Section of Preimage

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

== Theorem ==
Let $\struct {S, *}$ be a magma.
Let $\struct {T, \circ}$ be a magma with a zero element $0$.
Let $\phi: S \to T$ be a magma homomorphism.
Then $\struct {\phi^{-1} \sqbrk 0, *}$ is a submagma of $\struct {S, *}$.
== Proof ==
Let $x, y \in \phi^{-1} \sqbrk 0$.
It is to be shown that:
:$x * y \in \phi^{-1} \sqbrk 0$
Thus:
Hence the result.
Category:Abstract Algebra
Preimages All Exist iff Surjection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^{-1}$ be the inverse of $f$.
Let $\map {f^{-1} } t$ be the preimage of $t \in T$.
Then $\map {f^{-1} } t$ is empty for no $t \in T$ $f$ is a surjection.
=== Corollary ===
== Proof 1 ==
== Direct Proof ==
Category:Surjections
Category:Inverse Mappings
Category:Preimages All Exist iff Surjection
Preimages All Exist iff Surjection/Corollary

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^{-1}$ be the inverse of $f$.
:$\forall B \subseteq T, B \ne \O: f^{-1} \sqbrk B \ne \O$
:
:$f$ is a surjection
where $f^{-1} \sqbrk B$ denotes the preimage of $B \subseteq T$.
== Proof ==
=== Necessary Condition ===
Let $f$ be a surjection.
Let $B \subseteq T$ such that $B \ne \O$.
Then:
:$\exists t \in T: t \in B$
From Preimages All Exist iff Surjection:
:$\map {f^{-1} } t \ne \O$
As $t \in B$ it follows from Preimage of Subset is Subset of Preimage that:
:$f^{-1} \sqbrk B \ne \O$
$B$ is arbitrary, so:
:$\forall B \subseteq T, B \ne \O: f^{-1} \sqbrk B \ne \O$
=== Sufficient Condition ===
Suppose that:
:$\forall B \subseteq T, B \ne \O: f^{-1} \sqbrk B \ne \O$
 $f$ is not a surjection.
Then by definition:
:$\exists t \in T: \neg \paren {\exists s \in S: \map f s = t}$
That is:
:$\exists \set t \subseteq T: f^{-1} \sqbrk {\set t} = \O$
which contradicts the hypothesis.
So by Proof by Contradiction, $f$ is a surjection.
Hence the result.
== Sources ==
* : $\S 10$: Inverses and Composites
Category:Surjections
Category:Inverse Mappings
Preimages All Exist iff Surjection/Proof 1

== Theorem ==
== Proof ==
=== Necessary Condition ===
We use a Proof by Contraposition.
To that end, suppose:
:$\exists t \in T: \map {f^{-1} } t = \O$
That is:
:$\neg \paren {\forall t \in T: \exists s \in S: \map f s = t}$
So, by definition, $f: S \to T$ is not a surjection.
From Rule of Transposition it follows that if $f$ is a surjection:
$\neg \exists t \in T: \map {f^{-1} } t = \O$
=== Sufficient Condition ===
We again use a Proof by Contraposition.
To that end, suppose $f$ is not a surjection.
Then by definition:
:$\exists t \in T: \neg \paren {\exists s \in S: \map f s = t}$
That is:
:$\exists t \in T: \map {f^{-1} } t = \O$
From Rule of Transposition it follows that if $\neg \exists t \in T: \map {f^{-1} } t = \O$, then $f$ is a surjection.
Category:Preimages All Exist iff Surjection
Preimages All Exist iff Surjection/Proof 2

== Theorem ==
== Proof ==
Suppose that there is no $t \in T$ such that $\map {f^{-1} } t$ is empty.
By Denial of Existence, this is equivalent to saying that for all $t \in T$, $\map {f^{-1} } t$ is not empty.
This is equivalent to the statement that $\map {f^{-1} } t$ contains at least one element for each $t \in T$.
In other words, for each $t \in T$, there exists an $s\in S$ such that $\map f s = t$.
This is the definition of $f$ being surjective.
Thus if there is no $t \in T$ such that $\map {f^{-1} } t$ is empty, then $f$ is surjective.
Since this proof only uses statements of equivalence, it also shows that $f$ being surjective implies that there is no $t \in T$ such that $\map {f^{-1} } t$ is empty.
Category:Preimages All Exist iff Surjection