nyuuzyou/wikis huggingface dataset:
Preimage of Intersection under Relation/General Result

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\map \PP T$ be the power set of $T$.
Let $\mathbb T \subseteq \map \PP T$.
Then:
:$\ds \RR^{-1} \sqbrk{\bigcap \mathbb T} \subseteq \bigcap_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$
== Proof ==
This follows from Image of Intersection under Relation: General Result, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules.
Category:Relation Theory
Category:Set Intersection
Preimage of Lower Section under Increasing Mapping is Lower

== 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 lower section of $T$.
Then:
:$f^{-1} \sqbrk X$ is lower
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 $y \preceq x$.
Then:
== Sources ==
* 
Category:Lower Sections
Preimage of Mapping equals Domain

== Theorem ==
The preimage of a mapping is the same set as its domain:
:$\Preimg f = \Dom f$
== Proof ==
Let $f \subseteq S \times T$ be a mapping.
Then:
From Preimage of Relation is Subset of Domain, we have that $\Preimg f \subseteq S$.
The result follows from the definition of set equality.
Category:Preimages under Mappings
Category:Domains of Mappings
Preimage of Maximal Ideal of Finitely Generated Algebra is Maximal

== Theorem ==
Let $k$ be a field.
Let $A$ and $B$ be $k$-algebras.
Let $f: A \to B$ be a $k$-algebra homomorphism.
Let $B$ be finitely generated over $k$.
Let $\mathfrak m$ be a maximal ideal of $B$.
Then its preimage $\map {f^{-1} } {\mathfrak m}$ is a maximal ideal of $A$.
== Proof ==
We have an injective morphism:
:$\dfrac A {\map {f^{-1} } {\mathfrak m} } \to \dfrac B {\mathfrak m}$
We have that $\dfrac B {\mathfrak m}$ is a field extension of $k$ which is finitely generated.
Thus, by Zariski's Lemma, $\dfrac B {\mathfrak m}$ is a finite field extension. 
By Subalgebra of Finite Field Extension is Field, $\dfrac A {\map {f^{-1} } {\mathfrak m} }$ is a field.
Thus $\map {f^{-1} } {\mathfrak m}$ is a maximal ideal.
== Also see ==
* Definition:Spectrum of k-Algebra Functor
Category:Commutative Algebra
Category:Algebraic Geometry
Preimage of Maximum of Bounded Linear Functional on Extreme Set in Convex Compact Set is Extreme Set

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Let $K$ be a compact convex subset of $X$. 
Let $M \subseteq K$ be an extreme set in $K$. 
Let $f \in X^\ast$. 
Let:
:$\ds M^f = \set {x \in M : \map f x = \max_{y \in M} \map f y}$
Then $M^f$ is an extreme set in $K$.
== Proof ==
Note that from the extreme value theorem for normed vector spaces we have that: 
:$f$ attains its supremum
and so $M^f$ is well-defined.
From Mapping is Continuous iff Inverse Images of Open Sets are Open: Corollary, we have: 
:$M^f$ is closed.
Let $x, y \in K$ and $t \in \openint 0 1$ be such that: 
:$t x + \paren {1 - t} y \in M^f$
We show that $x, y \in M^f$. 
Since $M^f \subseteq M$, we have: 
:$t x + \paren {1 - t} y \in M$
and so $x, y \in M$ since $M$ is an extreme set in $K$. 
Since $t x + \paren {1 - t} y \in M^f$, we have: 
:$\ds \map f {t x + \paren {1 - t} y} = \max_{z \in M} \map f z$
Since $f \in X^\ast$, we have: 
:$\ds t \map f x + \paren {1 - t} \map f y = \max_{z \in M} \map f z$
On the other hand, since $x, y \in M$, we have: 
:$\ds \map f x \le \max_{z \in M} \map f z$
and:
:$\ds \map f y \le \max_{z \in M} \map f z$ 
from the definition of maximum.
If: 
:$\ds \map f x < \max_{z \in M} \map f z$
or: 
:$\ds \map f y < \max_{z \in M} \map f z$
we have: 
:$\ds t \map f x + \paren {1 - t} \map f y < \max_{z \in M} \map f z$
So, we must have: 
:$\ds \map f x = \map f y = \max_{z \in M} \map f z$
So:
:$x, y \in M^f$
So:
:$M^f$ is an extreme set in $K$.
== Sources ==
* $21.6$: The Krein-Milman Theorem
Category:Extreme Sets
Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal

== Theorem ==
Let $G$ be a group.
Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$.
Let $K \lhd G / H$.
Let $L = q_H^{-1} \sqbrk K$, where:
:$q_H: G \to G / H$ is the quotient epimorphism from $G$ to the quotient group $G / H$
:$q_H^{-1} \sqbrk K$ is the preimage of $K$ under $q_H$.
Then:
:$L \lhd G$
and there exists a group isomorphism $\phi: \paren {G / H} / K \to G / L$ defined as:
:$\phi \circ q_K \circ q_H = q_L$
== Proof 1 ==
== Proof 2 ==
Category:Group Isomorphisms
Category:Group Epimorphisms
Category:Quotient Groups
Category:Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal
Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal/Proof 1

== Theorem ==
== Proof ==
By Quotient Mapping on Structure is Epimorphism, both $q_K$ and $q_H$ are epimorphisms.
From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \paren {G / H} / K$ is also an epimorphism.
Now:
:$\forall x \in G: x \in \map \ker {q_K \circ q_H} \iff \map {q_K} {\map {q_H} x} = K = e_{G / H}$
This means the same as:
:$\map {q_H} x \in \map \ker {q_K} = K$
But:
:$\map {q_H} x \in K \iff x \in \map {q_H^{-1} } K = L$
Thus:
:$L = \map \ker {q_K \circ q_H}$
By Kernel is Normal Subgroup of Domain:
:$L \lhd G$
By Quotient Mapping on Structure is Epimorphism, both $q_K$ and $q_H$ are epimorphisms.
From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \paren {G / H} / K$ is also an epimorphism.
Then :
:$L \lhd G$
Hence by Quotient Theorem for Group Epimorphisms:
:there exists a group isomorphism $\psi: G / L \to \paren {G / H} / K$ satisfying:
::$\psi \circ q_L = q_K \circ q_L$
Let $\phi = \psi^{-1}$.
Then $\phi$ is a group isomorphism from $\paren {G / H} / K$ to $G / L$:
:$\phi \circ q_k \circ q_H = \phi \circ \psi \circ q_L = q_L$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Example $12.4$
Category:Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal
Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal/Proof 2

== Theorem ==
== Proof ==
Let $e$ be the identity element of $G$.
Let $\RR$ be the congruence relation defined by $H$ in $G$.
Let $\SS$ be the congruence relation defined by $K$ in $G / H$.
Let $\TT$ be the relation on $G$ defined as:
:$\forall x, y \in G: x \mathrel \TT y \iff x H \mathrel \SS y H$
From Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence:
:$\TT$ is a congruence relation on $G$
Hence from Congruence Relation on Group induces Normal Subgroup:
:the equivalence class under $\TT$ of $e$, that is $\eqclass e \TT$, is a normal subgroup of $G$.
Then we have:
Recall that $H$ is the identity of $G / H$.
Then as $K$ is a subgroup of $G / H$:
:$H \in K$
from Identity of Subgroup.
Then:
That is:
:$\eqclass e \TT = L$
and so $L$ is a normal subgroup of $G$ .
We can identify:
From Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence we have:
:there exists a unique isomorphism $\phi$ from $\paren {G / H} / K$ to $G / L$ which satisfies:
::$\phi \circ q_\SS \circ q_\RR = q_\TT$
Thus using the above identfications:
:$\phi \circ q_K \circ q_H = q_L$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.17 \ \text {(b)}$
Category:Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal
Preimage of Normal Subgroup under Epimorphism is Normal Subgroup

== 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 $H$ be a normal subgroup of $\struct {G_2, *}$.
Then:
:$\phi^{-1} \sqbrk H$ is a normal subgroup of $\struct {G_1, \circ}$
where $\phi^{-1} \sqbrk H$ denotes the preimage of $H$ under $\phi$.
== Proof ==
Recall that as $\phi$ is an group epimorphism, it is also a group Homomorphism.
Hence:
:$\forall a, b \in G_1: \map \phi {a \circ b} = \map \phi a * \map \phi b$
Let $H$ be a normal subgroup of $\struct {G_2, *}$.
From Preimage of Subgroup under Epimorphism is Subgroup:
:$\phi^{-1} \sqbrk H$ is a subgroup of $\struct {G_1, \circ}$.
It remains to be shown that $\phi^{-1} \sqbrk H$ is normal.
Let $K = \phi^{-1} \sqbrk H$.
That is:
:$H = \map \phi K$
To prove that $K$ is normal, we are to show that:
:$\forall g \in G: k \in K \iff g \circ k \circ g^{-1} \in K$
and:
:$\forall g \in G: k \in K \iff g^{-1} \circ k \circ g \in K$
So, let $g \in G$ and $k \in K$ be arbitrary.
By definition of preimage of $H$ under $\phi$:
:$\exists h \in H: \map \phi k = h$
Then:
In the same way:
Thus:
:$\forall g \in G: k \in K \implies g \circ k \circ g^{-1} \in K$
and:
:$\forall g \in G: k \in K \implies g^{-1} \circ k \circ g \in K$
Then:
and:
Thus:
:$\forall g \in G: k \in K \impliedby g \circ k \circ g^{-1} \in K$
and:
:$\forall g \in G: k \in K \impliedby g^{-1} \circ k \circ g \in K$
Putting this together:
:$\forall g \in G: k \in K \iff g \circ k \circ g^{-1} \in K$
and:
:$\forall g \in G: k \in K \iff g^{-1} \circ k \circ g \in K$
and the result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(b)}$
Category:Group Epimorphisms
Category:Normal Subgroups
Preimage of Open Sets forms Basis if Continuous Mappings Separate Points from Closed Sets

== Theorem ==
Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\family {f_i}_{i \mathop \in I}$ separate points from closed sets.
Let $\BB = \set{f_i^{-1} \sqbrk V : i \in I, V \text{ is open in } Y_i}$.
Then:
:$\BB$ is a basis for $X$
== Proof ==
By definition of continuous:
:$\forall B \in \BB : B$ is open in $X$
Let $U \subseteq X$ be open in $X$.
Let $x \in U$.
By definition of closed subset:
:$X \setminus U$ is closed in $X$
By definition of mappings separating points from closed sets:
:$\exists i \in I : \map {f_i} x \notin f_i \sqbrk {X \setminus U}^-$
Hence:
:$\map {f_i} x \in Y_i \setminus f_i \sqbrk {X \setminus U}^-$
By definition of preimage of mapping:
:$x \in f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-}$
From Topological Closure is Closed:
:$f_i \sqbrk {X \setminus U}^-$ is closed in $Y_i$
By definition of closed subset:
:$Y_i \setminus f_i \sqbrk {X \setminus U}^-$ is open in $Y_i$
Hence:
:$f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-} \in \BB$
We have:
By definition of subset:
:$f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-} \subseteq U$
We have shown that:
:$\forall$ open subsets $U \subseteq X, x \in U : \exists B \in \BB : x \in B \subseteq U$
Hence by definition $\BB$ is a basis for $X$.
Category:Continuous Mappings
Preimage of Prime Ideal under Ring Homomorphism is Prime Ideal

== Theorem ==
Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak p \subseteq B$ be a prime ideal.
Then its preimage $\map {f^{-1} } {\mathfrak p}$ is a prime ideal of $A$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Definition:Induced Mapping on Spectra of Rings
Category:Prime Ideals of Rings
Category:Preimage of Prime Ideal under Ring Homomorphism is Prime Ideal
Preimage of Prime Ideal under Ring Homomorphism is Prime Ideal/Proof 1

== Theorem ==
== Proof ==
By Preimage of Ideal under Ring Homomorphism is Ideal, $\map {f^{-1} } {\mathfrak p}$ is a ideal of $A$.
Let $a, b \in A$ with $a b \in \map {f^{-1} } {\mathfrak p}$.
By definitin of preimage, $\map f a \, \map f b = \map f {a b} \in \mathfrak p$.
Because $\mathfrak p$ is prime, $\map f a \in \mathfrak p$ or $\map f b \in \mathfrak p$.
Thus $a \in \map {f^{-1} } {\mathfrak p}$ or $b \in \map {f^{-1} } {\mathfrak p}$.
Thus $\map {f^{-1} } {\mathfrak p}$ is a prime ideal.
Category:Preimage of Prime Ideal under Ring Homomorphism is Prime Ideal
Preimage of Prime Ideal under Ring Homomorphism is Prime Ideal/Proof 2

== Theorem ==
== Proof ==
By Preimage of Ideal under Ring Homomorphism is Ideal, $\map {f^{-1} } {\mathfrak p}$ is a ideal of $A$.
Let $B / \mathfrak p$ be the quotient ring.
By Prime Ideal iff Quotient Ring is Integral Domain, $B / \mathfrak p$ is an integral domain.
By Quotient Ring of Kernel of Ring Epimorphism, the quotient ring epimorphism $B \to B / \mathfrak p$ has kernel $\mathfrak p$.
By Kernel of Composition of Ring Homomorphisms, the composition $A \overset f \to B \to B / \mathfrak p$ has kernel $\map {f^{-1} } {\mathfrak p}$.
By Universal Property of Quotient Ring, there exists a ring homomorphism $A / \map {f^{-1} } {\mathfrak p} \to B / \mathfrak p$.
.
By Subring of Integral Domain is Integral Domain, $A / \map {f^{-1} } {\mathfrak p}$ is an integral domain.
By Prime Ideal iff Quotient Ring is Integral Domain, $\map {f^{-1} } {\mathfrak p}$ is a prime ideal of $A$.
Category:Preimage of Prime Ideal under Ring Homomorphism is Prime Ideal
Preimage of Relation is Subset of Domain

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Then the preimage of $\RR$ is a subset of its domain:
:$\Preimg \RR \subseteq S$
== Proof ==
The preimage of $\RR$ is defined as:
:$\Preimg \RR = \set {s \in \Dom \RR: \exists t \in \Rng \RR: \tuple {s, t} \in \RR}$
Hence:
Category:Relation Theory
Preimage of Set Difference under Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \setminus T_2} = f^{-1} \sqbrk {T_1} \setminus f^{-1} \sqbrk {T_2}$
where:
:$\setminus$ denotes set difference
:$f^{-1} \sqbrk {T_1}$ denotes preimage.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
From Inverse of Mapping is One-to-Many Relation, we have that $f^{-1}: T \to S$ is one-to-many.
Thus we can apply One-to-Many Image of Set Difference:
:$\RR \sqbrk {T_1 \setminus T_2} = \RR \sqbrk {T_1} \setminus \RR \sqbrk {T_2}$
where in this context $\RR = f^{-1}$.
== Sources ==
* : Chapter $0$: Functions: Theorem $7 \ \text{(a)}$
* : Chapter $1$: Mappings: $\S 12 \alpha$
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Exercise $4 \ \text{(b)}$
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.3 \ \text{(iii)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(d)}$
* : $\S 2$
Category:Preimages under Mappings
Category:Set Difference
Preimage of Set Difference under Mapping/Corollary 1

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $T_1 \subseteq T_2 \subseteq T$.
Then:
:$\relcomp {f^{-1} \sqbrk {T_2} } {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp {T_2} {T_1} }$
where:
:$\complement$ (in this context) denotes relative complement
:$f^{-1} \sqbrk {T_1}$ denotes preimage.
== Proof ==
From One-to-Many Image of Set Difference: Corollary 1 we have:
:$\relcomp {\RR \sqbrk {T_2} } {\RR \sqbrk {T_1} } = \RR \sqbrk {\relcomp {T_2} {T_1} }$
where $\RR \subseteq T \times S$ is a one-to-many relation on $T \times S$.
Hence as $f^{-1}: T \to S$ is a one-to-many relation:
:$\relcomp {f^{-1} \sqbrk {T_2} } {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp {T_2} {T_1} }$
Category:Preimages under Mappings
Category:Set Difference
Category:Set Complement
Preimage of Set Difference under Relation

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Let $C$ and $D$ be subsets of $T$.
Then:
:$\RR^{-1} \sqbrk C \setminus \RR^{-1} \sqbrk D \subseteq \RR^{-1} \sqbrk {C \setminus D}$
where:
:$\setminus$ denotes set difference
:$\RR^{-1} \sqbrk C$ denotes the preimage of $C$ under $\RR$.
== Proof ==
We have that $\RR^{-1}$ is itself a relation
The result then follows from Image of Set Difference under Relation.
Category:Preimages under Relations
Category:Set Difference
Preimage of Singleton

== Theorem ==
Let $\RR$ be a relation.
Let $\map {\RR^{-1} } t$ denote the preimage of $t$ under $\RR$.
Let $\RR^{-1} \sqbrk {\set t}$ denote the preimage of $\set t$ under $\RR$.
Then:
:$\RR^{-1} \sqbrk {\set t} = \map {\RR^{-1} } t$
== Proof ==
== Sources ==
* : $\S 6.19$
Category:Relation Theory
Category:Singletons
Preimage of Subgroup under Epimorphism is Subgroup

== 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 $H$ be a subgroup of $\struct {G_2, *}$.
Then:
:$\phi^{-1} \sqbrk H$ is a subgroup of $\struct {G_1, \circ}$
where $\phi^{-1} \sqbrk H$ denotes the preimage of $H$ under $\phi$.
== Proof ==
Let $H$ be a subgroup of $\struct {G_2, *}$.
First note that from Null Relation is Mapping iff Domain is Empty Set:
:$\phi^{-1} \sqbrk H = \O \implies H = \O$
But $H \ne \O$.
Hence $\phi^{-1} \sqbrk H$ is not empty.
Next, let $x, y \in \phi^{-1} \sqbrk H$.
Then:
:$\exists h_1, h_2 \in H: h_1 = \map \phi x, h_2 = \map \phi y$
From the definition of Group Homomorphism, we have:
:$\map \phi {x^{-1} \circ y} = h_1^{-1} * h_2$
Since $H$ is a subgroup:
:$h_1^{-1} * h_2 \in H$
Hence:
:$x^{-1} \circ y \in \phi^{-1} \sqbrk H$
Thus from the One-Step Subgroup Test:
:$\phi^{-1} \sqbrk H$ is a subgroup of $G_1$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(b)}$
Category:Group Epimorphisms
Category:Subgroups
Preimage 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 $N$ be a submodule of $H$.
Then $\phi^{-1} \sqbrk N$ is a submodule of $G$.
== Proof ==
Let $M = \phi^{-1} \sqbrk N$ be the preimage of $N$ under $\phi$.
 $M$ is not a submodule of $G$.
This means that $M$ does not fulfil all the module axioms.
First suppose that:
:$(1): \quad \struct {M, +_G}$ is not a subgroup of $G$.
Then $M$ is not a group.
Then by the One-Step Subgroup Test:
:$\exists x, y \in M: x +_G \paren {-y} \notin M$
Then:
But by definition of $\phi$, both $\map \phi x \in N$ and $\map \phi {-y} \in N$.
Hence $\struct {N, +_G}$ is not closed.
By as applied to $N$, it follows that $N$ is not a group.
Hence $N$ is not a subgroup of $H$.
From this contradiction it follows that $(1)$ is false.
Thus $\struct {M, +_G}$ is a subgroup of $G$.
We have that $M$ is not a submodule of $G$.
We have shown that $\struct {M, +_G}$ is a subgroup of $G$.
Hence it must be the case that $M$ is not closed for scalar product:
But by definition of $\phi$, both $\map \phi x \in N$.
Hence $N$ is not closed for scalar product.
From this contradiction it follows that $(2)$ is false.
Thus $\struct {M, +_G}$ is closed for scalar product.
We have shown that $M$ is a subgroup of $G$ which is closed for scalar product.
That is, $M$ is a submodule of $G$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.2$
Category:Linear Transformations
Preimage of Subring under Ring Homomorphism is Subring

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S_2$ be a subring of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$.
== Proof ==
Let $K = \map \ker \phi$ be the kernel of $R_1$.
We have that $0_{R_2} \in S_2$ and so $\set {0_{R_2} } \subseteq S_2$.
From Subset Maps to Subset:
:$\phi^{-1} \sqbrk {\set {0_{R_2} } } \subseteq \phi^{-1} \sqbrk {S_2} = S_1$
But by definition, $K = \phi^{-1} \sqbrk {\set {0_{R_2} } }$
and so $S_1$ is a subset of $R_1$ containing $K$, that is:
:$K \subseteq S_1 \subseteq R_1$
Now we need to show that $S_1$ is a subring of $R_1$.
Let $r, r' \in S_1$.
Then $\map \phi r, \map \phi {r'} \in S_2$.
Hence:
So:
:$r + r' \in S_1$
Then:
So:
:$-r \in \phi^{-1} S_1$
Finally:
So:
:$r \circ_1 r' \in S_1$
So from the Subring Test we have that $\phi^{-1} \sqbrk {S_2}$ is a subring of $R$ containing $K$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 3^\circ$
Category:Ring Homomorphisms
Category:Subrings
Preimage of Subset is Subset of Preimage

== Corollary to Image of Subset under Relation is Subset of Image ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $C, D \subseteq T$.
Then:
:$C \subseteq D \implies f^{-1} \sqbrk C \subseteq f^{-1} \sqbrk D$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall C, D \in \powerset T: C \subseteq D \implies \map {f^\gets} C \subseteq \map {f^\gets} D$
== Proof ==
As $f: S \to T$ is a mapping, it is also a relation, and thus so is its inverse:
:$f^{-1} \subseteq T \times S$
The result follows directly from Image of Subset under Relation is Subset of Image.
== Sources ==
* : $\S 2.2$: Homomorphisms: $\text{(ii)}$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(j)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(a)}$
Category:Subsets
Category:Preimages under Mappings
Preimage of Subset of Cartesian Product under Injection from Factor

== Theorem ==
Let $S$ and $T$ be sets such that $T \ne \O$.
Let $S \times T$ denote their cartesian product.
Let $t \in T$ be given.
Let $j_t: S \to S \times T$ be the injection from $S$ to $S \times T$ defined as:
:$\forall s \in S: \map {j_t} s = \tuple {s, t}$
Let $W \subseteq S \times T$.
Let $V = {j_t}^{-1} \sqbrk W$ denote the preimage of $W$ under $j_t$.
Then:
:$V = \set {s \in S : \tuple {s, t} \in W}$
== Proof ==
That $j_t$ is actually an injection is demonstrated in Mapping from Set to Ordinate of Cartesian Product is Injection.
Then:
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $6$
Category:Cartesian Product
Category:Injections
Preimage of Subset under Composite Mapping

== Theorem ==
Let $S_1, S_2, S_3$ be sets.
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings.
Denote with $g \circ f: S_1 \to S_3$ the composition of $g$ and $f$.
Let $S_3' \subseteq S_3$ be a subset of $S_3$.
Then:
:$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$
where $g^{-1} \sqbrk {S_3'}$ denotes the preimage of $S_3'$ under $g$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 8$: Composition of Functions and Diagrams: Exercise $4$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.4 \ \text{(a)}$
Category:Composite Mappings‎
Category:Preimages under Mappings
Category:Preimage of Subset under Composite Mapping
Preimage of Subset under Composite Mapping/Proof 1

== Theorem ==
== Proof ==
A mapping is a specific kind of relation.
Hence, Inverse of Composite Relation applies, and it follows that:
:$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$
Category:Preimage of Subset under Composite Mapping
Preimage of Subset under Identity Mapping

== Theorem ==
Let $S$ be a set.
Let $\iota_S: S \to S$ be the identity mapping on $S$.
Let $T \subseteq S$.
Then:
:$\iota_S^{-1} \sqbrk T = T$
where $\iota_S^{-1} \sqbrk T$ is the preimage of $T$ under $\iota_S$.
== Proof ==
By definition of identity mapping:
:$\iota_S: S \to S: \forall x \in S: \map {\iota_S} x = x$
Let $i_S: S \to S$ be the inclusion mapping of $S$ into $S$.
By definition of inclusion mapping:
:$i_S: S \to S: \forall x \in S: \map {i_S} x = x$
From Equality of Mappings:
:$\iota_S = i_S$
From Preimage of Subset under Inclusion Mapping:
:$\forall T\subseteq S : \iota_S^{-1} \sqbrk T = T \cap S$
From Intersection with Subset is Subset:
:$\forall T\subseteq S : T \cap S = T$
Hence:
:$\forall T\subseteq S : \iota_S^{-1} \sqbrk T = T$
Category:Identity Mappings
Category:Subsets