nyuuzyou/wikis huggingface dataset:
Image of Preimage under Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B = B \cap f \sqbrk S$
where:
:$f \sqbrk S$ denotes the image of $S$ under $f$
:$f^{-1} \sqbrk B$ denotes the preimage of $B$ under $f$
:$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.
=== Corollary ===
== Proof ==
As $f$ is a mapping it follows by definition that $f$ is also a relation
Applying Image of Preimage under Relation is Subset directly:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B \implies f \sqbrk {f^{-1} \sqbrk B} \subseteq B$
From Preimage of Subset is Subset of Preimage:
:$B \subseteq T \implies f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk T$
From Preimage of Mapping equals Domain:
:$f^{-1} \sqbrk T = S$
and so:
:$B \subseteq T \implies f^{-1} \sqbrk B \subseteq S$
Applying Image of Subset under Mapping is Subset of Image:
:$f^{-1} \sqbrk B \subseteq S \implies f \sqbrk {f^{-1} \sqbrk B} \subseteq f \sqbrk S$
From Intersection is Largest Subset:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B \cap f \sqbrk S$
Now suppose $y \in B \cap f \sqbrk S$.
Then:
:$f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk B$ and $f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk {f \sqbrk S}$
and in particular:
:$f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk B$
Applying Image of Subset under Mapping is Subset of Image again:
:$f \sqbrk {f^{-1} \sqbrk y} \subseteq f \sqbrk {f^{-1} \sqbrk B}$
But as $f$ is many-to-one:
:$f \sqbrk {f^{-1} \sqbrk y} = y$
and so:
:$y \in f \sqbrk {f^{-1} \sqbrk B} = \paren {f \circ f^{-1} } \sqbrk B$
So we have that:
:$B \cap f \sqbrk S \subseteq \paren {f \circ f^{-1} } \sqbrk B$
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(d)}$
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.7 \ \text{(d)}$
Category:Preimages under Mappings
Image of Preimage under Mapping/Corollary

== Corollary to Image of Preimage under Mapping ==
Let $f: S \to T$ be a mapping.
Then:
:$B \subseteq \Img f \implies \paren {f \circ f^{-1} } \sqbrk B = B$
where:
:$f \sqbrk X$ denotes the image of $X$ under $f$
:$f^{-1} \sqbrk X$ denotes the preimage of $X$ under $f$
:$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$
:$\Img f$ denotes the image set of $f$.
== Proof ==
From Preimage of Subset is Subset of Preimage:
:$B \subseteq \Img f \implies f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img f}$
and from Intersection with Subset is Subset:
:$f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img f} \implies f^{-1} \sqbrk B \cap f^{-1} \sqbrk {\Img f} = f^{-1} \sqbrk B$
Hence the result.
== Sources ==
* : $\S 2.2$: Homomorphisms: $\text{(i)}$
Category:Preimages under Mappings
Image of Preimage under Relation is Subset

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$B \subseteq T \implies \paren {\RR \circ \RR^{-1} } \sqbrk B \subseteq B$
where:
:$\RR \sqbrk B$ denotes the image of $B$ under $\RR$
:$\RR^{-1} \sqbrk B$ denotes the preimage of $B$ under $\RR$
:$\RR \circ \RR^{-1}$ denotes composition of $\RR$ and $\RR^{-1}$.
== Proof ==
Let $B \subseteq T$.
Then:
So by definition of subset:
:$B \subseteq T \implies \paren {\RR \circ \RR^{-1} }\sqbrk B \subseteq B$
== Also see ==
* Subset of Domain is Subset of Preimage of Image
Category:Preimages under Relations
Category:Composite Relations
Image of Projection in Plane

== Theorem ==
Let $M$ and $N$ be distinct lines in the plane.
:
Let $\pr_{M, N}$ be the projection on $M$ along $N$:
:$\forall x \in \R^2: \map {\pr_{M, N} } x =$ the intersection of $M$ with the line through $x$ parallel to $N$.
Then $M$ is the image of $\pr_{M, N}$.
== Proof ==
Let $x \in \R^2$ be arbitrary.
By definition, the image of $x$ is the intersection of $M$ with the line through $x$ parallel to $N$.
Therefore $\map {\pr_{M, N} } x \in M$.
Hence:
:$\Img {\pr_{M, N} } \subseteq M$.
Now consider $y \in M$.
By Playfair's axiom there exists exactly one straight line $L$ parallel to $N$ passing through $y$.
Hence $y$ is the image of every point on $L$.
As $y$ is arbitrary, it follows that each point on $M$ is the image of some straight line in $\R^2$ under $\pr_{M, N}$.
That is:
:$M \subseteq \Img {\pr_{M, N} }$
The result follows by definition of set equality.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.5$
Category:Geometric Projections
Image of Proper Subset under Progressing Mapping on Minimally Closed Class

== Theorem ==
Let $N$ be a class which is closed under a progressing mapping $g$.
Let $b$ be an element of $N$ such that $N$ is minimally closed under $g$ with respect to $b$.
Then:
:$x \subset y \implies \map g x \subseteq y$
== Proof ==
From Minimally Closed Class under Progressing Mapping induces Nest, we have that $N$ is a nest in which:
:$\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$
Thus the corollary 1 of the Sandwich Principle applies directly.
Category:Minimally Closed Classes under Progressing Mapping
Image of Real Natural Logarithm

== Theorem ==
Let $\ln$ be the natural logarithm function on the real numbers.
Then the image of $\ln$ is the set of real numbers:
:$\Img \ln = \R$
== Proof ==
By definition of natural logarithm:
:$\ln^{-1} = \exp$
From Exponential Tends to Zero and Infinity:
:$\Dom \exp = \openint {-\infty} {+\infty}$
:$\Img \exp = \openint 0 {+\infty}$
From Exponential is Strictly Increasing:
:$\exp$ is strictly increasing.
From Strictly Monotone Real Function is Bijective, $\exp: \R \to \R_{>0}$ is a bijection.
Thus:
:$\Img \ln = \Dom \exp$
and so $\Img \ln = \R$.
Category:Natural Logarithms
Image of Real Square Function

== Theorem ==
The image of the real square function is the entire set of positive real numbers $\R_{\ge 0}$.
== Proof ==
From Square of Real Number is Non-Negative, the image of $f$ is $\R_{\ge 0}$.
From Positive Real has Real Square Root:
:$\forall x \in \R: \exists y \in \R: x^2 = y$
Hence the result by definition of image.
== Sources ==
* : $\S 2.2$: Functions
Category:Square Function
Image of Relation is Domain of Inverse Relation

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^{-1} \subseteq T \times S$ be the inverse of $\RR$.
Then:
:$\Img \RR = \Dom {\RR^{-1} }$
That is, the image of a relation is the domain of its inverse.
== Proof ==
By definition:
== Also see ==
* Domain of Relation is Image of Inverse Relation
== Sources ==
* : Chapter $0$: Relations
* : $\S 10$: Inverses and Composites
* : $\S 1.11$: Relations
Category:Relation Theory
Category:Inverse Relations
Image of Relation is Subclass of Union of Union of Relation

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\Img \RR$ denote the image of $\RR$.
Then:
:$\Img \RR \subseteq \map \bigcup {\bigcup \RR}$
where $\bigcup \RR$ denotes the union of $\RR$.
== Proof ==
But then:
Also:
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 8$ Relations: Lemma $8.1$: $(2)$
Category:Class Union
Image of Relative Complement under Bijection is Relative Complement of Image

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection.
Let $H \subseteq S$.
Let $f \sqbrk H = K$ be the image of $H$ under $f$.
Let $\relcomp S H$ denote the relative complement of $H$ in $S$.
Then:
:$f \sqbrk {\relcomp S H} = \relcomp T K$
== Proof ==
From Set with Relative Complement forms Partition, $\set {H \mid \relcomp S H}$ forms a partition of $S$.
The result follows from Bijection Preserves Set Partition.
Category:Relative Complement
Category:Bijections
Image of Set Difference under Injection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $S_1$ and $S_2$ be subsets of $S$.
Let $S_1 \setminus S_2$ denote the set difference between $S_1$ and $S_2$.
Then:
:$\forall S_1, S_2 \subseteq S: f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$
 $f$ is an injection.
== Proof ==
An injection is a type of one-to-one relation, and therefore also a one-to-many relation.
Therefore One-to-Many Image of Set Difference applies:
:$\RR \sqbrk {S_1} \setminus \RR \sqbrk {S_2} = \RR \sqbrk {S_1 \setminus S_2}$
 $\RR$ is one-to-many.
We have that $f$ is a mapping and therefore a many-to-one relation.
So $f$ is a one-to-many relation $f$ is also an injection.
It follows that:
:$\forall S_1, S_2 \subseteq S: f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$
 $f$ is an injection.
== Sources ==
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Exercise $4 \ \text{(a)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(h)}$
* : $2.2$
* : $\S 10$: Inverses and Composites: Exercise $\text{(iii)}$
Category:Set Difference
Category:Injections
Image of Set Difference under Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
The set difference of the images of two subsets of $S$ is a subset of the image of the set difference.
That is:
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
:$f \sqbrk {S_1} \setminus f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \setminus S_2}$
where $\setminus$ denotes set difference.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Difference of Images under Mapping not necessarily equal to Image of Difference: equality does not hold in general
* Image of Set Difference under Injection: equality holds for an injection
== Sources ==
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(h)}$
* : $\S 2$: Problem $4 \ \text{(i)}$
Category:Image of Set Difference under Mapping
Category:Images
Category:Set Difference
Image of Set Difference under Mapping/Corollary 1

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $S_1 \subseteq S_2 \subseteq S$.
Then:
:$\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$
where $\complement$ (in this context) denotes relative complement.
== Proof ==
From Image of Set Difference under Relation: Corollary 1 we have:
:$\relcomp {\RR \sqbrk {S_2} } {\RR \sqbrk {S_1} } \subseteq \RR \sqbrk {\relcomp {S_2} {S_1} }$
where $\RR \subseteq S \times T$ is a relation on $S \times T$.
As $f$, being a mapping, is also a relation, it follows directly that:
:$\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$
Category:Image of Set Difference under Mapping
Image of Set Difference under Mapping/Corollary 2

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $X$ be a subset of $S$.
Then:
:$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$
where:
:$\Img f$ denotes the image of $f$
:$\complement_{\Img f}$ denotes the complement relative to $\Img f$.
This can be expressed in the language and notation of direct image mappings as:
:$\forall X \in \powerset S: \relcomp {\Img f} {\map {f^\to} X} \subseteq \map {f^\to} {\relcomp S X}$
That is:
:$\forall X \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } X \subseteq \map {\paren {f^\to \circ \complement_S} } X$
where $\circ$ denotes composition of mappings.
== Proof ==
From Image of Set Difference under Relation: Corollary 2 we have:
:$\relcomp {\Img \RR} {\RR \sqbrk X} \subseteq \RR \sqbrk {\relcomp S X}$
where $\RR \subseteq S \times T$ is a relation on $S \times T$.
As $f$, being a mapping, is also a relation, it follows directly that:
:$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1 \ \text{(iv)}$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(c)}$
Category:Image of Set Difference under Mapping
Image of Set Difference under Mapping/Corollary 3

== Theorem ==
Let $f: S \to T$ be a surjection.
Let $A \subseteq S$ be a subset of $S$.
Then:
:$T \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$
where $\setminus$ denotes set difference.
== Proof ==
As $T$ is a surjection, $T = f \sqbrk S$.
Thus Image of Set Difference under Mapping can be applied:
:$f \sqbrk S \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$
== Sources ==
* : Chapter $4$: Mappings: Exercise $6$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(c)}$
Category:Image of Set Difference under Mapping
Image of Set Difference under Mapping/Proof 1

== Theorem ==
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Image of Set Difference under Relation:
:$\RR \sqbrk {S_1} \setminus \RR \sqbrk {S_2} \subseteq \RR \sqbrk {S_1 \setminus S_2}$
Category:Image of Set Difference under Mapping
Image of Set Difference under Relation

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Let $A$ and $B$ be subsets of $S$.
Then:
:$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$
where:
:$\setminus$ denotes set difference
:$\RR \sqbrk A$ denotes image of $A$ under $\RR$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
== Also see ==
Note that equality does not hold in general.
See Difference of Images under Mapping not necessarily equal to Image of Difference for an example of a mapping (which is of course a relation) for which it does not.
Category:Relation Theory
Category:Set Difference
Category:Image of Set Difference under Relation
Image of Set Difference under Relation/Corollary 1

== Corollary to Image of Set Difference under Relation ==
Let $\RR \subseteq S \times T$ be a relation.
Let $A \subseteq B \subseteq S$.
Then:
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp B A}$
where:
:$\RR \sqbrk B$ denotes the image of $B$ under $\RR$
:$\complement$ (in this context) denotes relative complement.
== Proof ==
We have that $A \subseteq B$.
Then by definition of relative complement:
:$\relcomp B A = B \setminus A$
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk B \setminus \RR \sqbrk A$
Hence, when $A \subseteq B$:
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp B A}$
means exactly the same thing as:
:$\RR \sqbrk B \setminus \RR \sqbrk A \subseteq \RR \sqbrk {B \setminus A}$
Category:Image of Set Difference under Relation
Image of Set Difference under Relation/Corollary 2

== Corollary to Image of Set Difference under Relation ==
Let $\RR \subseteq S \times T$ be a relation.
Let $A$ be a subset of $S$.
Then:
:$\relcomp {\Img \RR} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp S A}$
where:
:$\Img \RR$ denotes the image of $\RR$
:$\RR \sqbrk A$ denotes the image of $A$ under $\RR$.
== Proof ==
By definition of the image of $\RR$:
:$\Img \RR = \RR \sqbrk S$
So, when $B = S$ in Image of Set Difference under Relation: Corollary 1:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \relcomp {\RR \sqbrk S} {\RR \sqbrk A}$
Hence:
:$\relcomp {\Img \RR} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp S A}$
means exactly the same thing as:
:$\relcomp {\RR \sqbrk S} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp S A}$
that is:
:$\RR \sqbrk S \setminus \RR \sqbrk A \subseteq \RR \sqbrk {S \setminus A}$
Category:Image of Set Difference under Relation
Image of Set under Mapping is Set

== Theorem ==
Let $A$ be a class.
Let $\mathrm U$ denote the universal class.
Let $f: A \to \mathrm U$ be a class mapping.
Let $S$ be a subset of $A$.
Then the image $f \sqbrk S$ is also a set.
If $A$ is a set, then this result is known as the Axiom of Replacement in Zermelo-Fraenkel set theory.
== Proof ==
 $f \sqbrk S$ is not a set.
Then $f \sqbrk S$ must be proper.
By Restriction of Mapping to Image is Surjection, the restriction $f \restriction_{S \times f \sqbrk S}$ is a surjection.
But this contradicts Surjection from Class to Proper Class.
Thus by contradiction, $f \sqbrk S$ is a set.
Hence the result.
Category:Gödel-Bernays Class Theory
Category:Class Mappings
Image of Set under Mapping is Set iff Restriction is Set

== Theorem ==
Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $x$ be a set.
Let $f \sqbrk x$ denote the image of $x$ under $f$.
Let $f {\restriction} x$ denote the restriction of $f$ to $x$.
Then $f \sqbrk x$ is a set $f {\restriction} x$ is a set.
== Proof ==
Let $f {\restriction} x$ be a set.
Then its image is also a set.
But then:
:$\Img {f {\restriction} x} = f \sqbrk x$
Conversely let $f \sqbrk x$ be a set.
From Cartesian Product of Sets is Set:
:$x \times f \sqbrk x$ is a set.
From Restriction of Mapping is Subclass of Cartesian Product:
:$f {\restriction} A \subseteq A \times f \sqbrk A$
The result follows from Subclass of Set is Set.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Proposition $1.2$
Category:Images
Category:Restrictions
Image of Singleton under Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Then the image of an element of $S$ is equal to the image of a singleton containing that element, the singleton being a subset of $S$:
:$\forall s \in S: \set {\map f s} = f \sqbrk {\set s}$
== Proof ==
By definition, a mapping is a relation.
Thus Image of Singleton under Relation applies.
== Sources ==
* : $\S 1.10$: Functions: Remark $10.8 \ \text{(c)}$
Category:Images
Category:Singletons
Image of Singleton under Relation

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Then the image of an element of $S$ is equal to the image of a singleton containing that element, the singleton being a subset of $S$:
:$\forall s \in S: \map \RR s = \RR \sqbrk {\set s}$
== Proof ==
We have the definitions:
Finally:
:$\map \RR s = \RR \sqbrk {\set s}$
by definition of set equality.
Category:Relation Theory
Category:Singletons
Image of Small Class under Mapping is Small

== Theorem ==
Let $A$ be a mapping.
Let $a$ be a small class.
Then, the image of $a$ under $A$ is small.
== Proof ==
Since $A$ is a mapping:
:$\forall y: \exists x: \forall z: \left({ y A z \implies z = x }\right)$
This satisfies the antecedent of the axiom of replacement. Therefore:
:$\forall w: \exists x: \forall y: \left({ y \in w \implies \forall z: \left({ y A z \implies z \in x }\right) }\right)$
Universal Instantiation yields:
:$\exists x: \forall y: \left({ y \in a \implies \forall z: \left({ y A z \implies z \in x }\right) }\right)$
By applying the definition for the restricted universal quantifier and rearranging quantifiers:
:$\exists x: \forall z: \left({ \exists y \in a: y A z \implies z \in x }\right)$
Applying the definition for image:
:$\exists x: \operatorname{Im} \left({a}\right) \subseteq x$
By Axiom of Subsets Equivalents, the image of $a$ under $A$ must be small.
== Sources ==
* : $\S 6.7$
Category:Class Theory Category:Class Mappings
Image of Submodule under Linear Transformation is Submodule

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}_R$ and $\struct {H, +_H, \circ_H}_R$ be $R$-modules.
Let $\phi: G \to H$ be a linear transformation.
Let $M$ be a submodule of $G$.
Then $\phi \sqbrk M$ is a submodule of $H$.
== Proof ==
Let $N = \phi \sqbrk M$ be the image set of $M$ under $\phi$.
By definition, a linear transformation $\phi: G \to H$ is, in particular, a (group) homomorphism from the group $\struct {G, +_G}$ to the group $\struct {H, +_H}$.
We have that $M$ is a submodule of $G$.
So from Elements of Submodule form Subgroup, $M$ forms a subgroup of $G$.
From Group Homomorphism Preserves Subgroups, $N$ is therefore a subgroup of $H$.
It remains to be shown that $N$ is closed for scalar product:
:$\forall \lambda \in R, x \in N: \lambda \circ_H x \in N$
As $M$ is a submodule of $G$, $M$ itself is closed for scalar product:
:$\forall \lambda \in R, y \in N: \lambda \circ_G y \in M$
We have:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.2$
Category:Linear Transformations
Image of Subset is Image of Restriction

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $X \subseteq S$.
Let $f {\restriction_X}$ be the restriction of $f$ to $X$.
Then:
:$f \sqbrk X = \Img {f {\restriction_X} }$
where $\Img f$ denotes the image of $f$, defined as:
:$\Img f = \set {t \in T: \exists s \in S: t = \map f s}$
== Proof ==
Let $y \in f \sqbrk X$.
Then by definition of the image of a subset:
:$\exists x \in X: \map f x = y$
or equivalently:
:$\exists x \in X: \tuple {x, y} \in f$
But then by definition of restriction of $f$ to $X$:
:$f {\restriction_X} = \set {\tuple {x, y} \in f: x \in X}$
Thus by definition of the image set of $f {\restriction_X}$:
:$\Img {f {\restriction_X} } = f {\restriction_X} \sqbrk X = \set {y \in T: \exists x \in X: \tuple {x, y} \in f {\restriction_X} }$
Hence it can be seen that:
:$y \in \Img {f {\restriction_X} }$
So:
:$f \sqbrk X \subseteq \Img {f {\restriction_X} }$
Now suppose that $y \in \Img {f {\restriction_X} }$.
Then by definition of the image set of $f {\restriction_X}$:
:$\exists x \in S: \tuple {x, y} \in f {\restriction_X}$
By definition of restriction of $f$ to $X$:
:$x \in X$
Thus by definition of image of a subset:
:$y \in f \sqbrk X$
So:
:$\Img {f {\restriction_X} } \subseteq f \sqbrk X$
We have:
:$f \sqbrk X \subseteq \Img {f {\restriction_X} }$
:$\Img {f {\restriction_X} } \subseteq f \sqbrk X$
The result follows by definition of set equality.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.4 \ \text{(i)}$
Category:Restrictions
Image of Subset under 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 $\RR_2$.
Let $A \subseteq S_1$.
Then:
:$\RR_2 \sqbrk {\RR_1 \sqbrk A \cap S_2} = \paren{\RR_2 \circ \RR_1} \sqbrk A$
== Proof ==
We have:
== Also see ==
* Image of Element under Composite Relation
* Image of Subset under Composite Relation with Common Codomain and Domain
* Image of Element under Composite Relation with Common Codomain and Domain
Category:Composite Relations
Image of Subset under Composite Relation with Common Codomain and Domain

== Theorem ==
Let $\RR_1 \subseteq S \times T$ and $\RR_2 \subseteq T \times U$ be relations.
Let $\RR_2 \circ \RR_1 \subseteq S \times U$ be the composition of $\RR_1$ and $\RR_2$.
Let $A \subseteq S$.
Then:
:$\RR_2 \sqbrk {\RR_1 \sqbrk A} = \paren{\RR_2 \circ \RR_1} \sqbrk A$
== Proof ==
We have:
== Also see ==
* Image of Element under Composite Relation with Common Codomain and Domain
* Image of Subset under Composite Relation
* Image of Element under Composite Relation
Category:Composite Relations
Image of Subset under Mapping/Examples

== Examples of Images of Subsets under Mappings ==
=== Aribtrary Mapping from $\set {0, 1, 2, 3, 4, 5}$ to $\set {0, 1, 2, 3}$ ===
=== Image of $\closedint {-3} 2$ under $x \mapsto x^4 - 1$ ===
=== Image of $\closedint 1 2$ under $\map f x = x^2 - x - 2$ ===
=== Image of $\openint {-1} 1$ under $\map f x = x^2 - x - 2$ ===
Category:Examples of Images under Mappings
Image of Subset under Mapping/Examples/Arbitrary Finite Set

== Example of Images of Subsets under Mapping ==
Let:
Let $f: S \to S$ be the mapping defined as:
Let:
Then:
and:
:$\Img f = \set {0, 1, 3}$
== Sources ==
* : $\S 21.1$: The image of a subset of the domain; surjections
Category:Examples of Images under Mappings
Image of Subset under Mapping/Examples/Image of -1 to 1 under x^2-x-2

== Example of Image 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 image of the open interval $\openint {-1} 1$ is:
:$f \sqbrk {\openint {-1} 1} = \hointr {-\dfrac 9 4} 0$
== 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 $\map f {\dfrac 1 2} = -\dfrac 9 4$.
As $\dfrac 1 2 \in \closedint {-1} 1$, it can be seen that the minimum of $f \sqbrk {\openint {-1} 1}$ is $-\dfrac 9 4$.
As $f$ is strictly increasing on $x > 0$ and strictly decreasing on $x < 0$, it suffices to inspect the images of the endpoints $-1$ and $1$.
Thus:
:$\map f {-1} = \paren {-1}^2 - \paren {-1} - 2 = 0$
:$\map f 1 = 1^2 - 1 - 2 = -2$
The result follows.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions
Category:Examples of Images under Mappings