nyuuzyou/wikis huggingface dataset:
Image of Interval by Continuous Function is Interval/Proof 2

== Theorem ==
== Proof ==
Let $J$ be the image of $f$.
By Subset of Real Numbers is Interval iff Connected we need to show that $J$ is connected (and hence an interval).
 not.
Then there exists a separation $S \mid T$ of $J$.
Define $A = f^{-1} \sqbrk S$ and $B = f^{-1} \sqbrk T$. $A$ and $B$ are both non-empty.
Because $f$ is continuous, by Continuous iff inverse image of any open set is open we must have $A$ and $B$ open.
Because $S \mid T$ is a separation:
:$A \cap B = f^{-1} \sqbrk S \cap f^{-1} \sqbrk T = f^{-1} \sqbrk {S \cap T} = \O$, because $S \mid T$ is a separation.
Also, $A \cup B = f^{-1} \sqbrk S \cup f^{-1} \sqbrk T = f^{-1} \sqbrk {S \cup T} = f^{-1}(J) = I$ ($S \mid T$ is a separation of $J$).
Hence $A \mid B$ is a separation of $I$.
$I$ can certainly not be an interval (because it is not connected).
This is a contradiction.
Thus $J$ must be an interval.
== Sources ==
* : $2.7$
Category:Image of Interval by Continuous Function is Interval
Image of Interval by Derivative

== Theorem ==
Let $f$ be a real function that is everywhere differentiable.
Let $I \subseteq \Dom f$ be a real interval.
Then:
:$f' \sqbrk I$ is a real interval
where $f'$ denotes the derivative of $f$
== Proof ==
Let $x_1, x_2 \in f' \sqbrk I: x_1 < x_2$.
Let $\xi \in \openint {x_1} {x_2}$.
We need to show that $\xi \in f' \sqbrk I$.
Let $a, b \in I : \map {f'} a = x_1 \land \map {f'} b = x_2$.
, assume $a < b$. The case $b < a$ is handled similarly.
Let $\map g x = \map f x - \xi x$.
Then:
:$\map {g'} x = \map {f'} x - \xi$
By Differentiable Function is Continuous, $g$ is continuous.
By Restriction of Continuous Mapping is Continuous, $g \restriction_{\closedint a b}$ is continuous.
From Corollary 3 to Continuous Image of Compact Space is Compact:
:$g \restriction_{\closedint a b}$ attains a minimum.
By hypothesis, $\map {g'} a < 0$ and $\map {g'} b > 0$.
Let $\map {N^*_\epsilon} x$ denote the deleted $\epsilon$-neighborhood of $x$.
Then from Behaviour of Function Near Limit:
:$\exists \map {N^*_\epsilon} a: \forall x \in \map {N^*_\epsilon} a: \dfrac {\map g x - \map g a} {x - a} < 0$
and:
:$\exists \map {N^*_\delta} b: \forall x \in \map {N^*_\delta} b: \dfrac {\map g x - \map g b} {x - b} > 0$
Thus:
:$\exists x \in \map {N^*_\epsilon} a \cap \closedint a b: \map g x < \map g a$
and
:$\exists x \in \map {N^*_\delta} b \cap \closedint a b: \map g x < \map g b$
Hence $\map g a$ and $\map g b$ are not minima of $g \restriction_{\closedint a b}$.
So $g \restriction_{\closedint a b}$ must attain its minimum at some $m \in \openint a b$.
By Derivative at Maximum or Minimum:
:$\map {g'} m = 0$
Hence:
:$\map {f'} m = \xi$
The result follows.
== Sources ==
* : $\S 5.2.1$
Category:Real Intervals
Category:Differential Calculus
Image of Inverse Image

== Theorem ==
Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Let $X$ be a subset of $T$.
Then:
:$f \sqbrk {f^{-1} \sqbrk X} \subseteq X$
where:
:$f^{-1} \sqbrk X$ denotes the image of $X$ under the relation $f^{-1}$.
== Proof ==
Let $x \in f \sqbrk {f^{-1} \sqbrk X}$.
By definition of image of set:
:$\exists y \in S: y \in f^{-1} \sqbrk X \land x = \map f y$
By definition of image of set under relation:
:$\map f y \in X$
Thus $x \in X$
== Sources ==
* 
Category:Images
Category:Inverse Mappings
Image of Linear Combination of Subsets of Vector Space under 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 $\lambda, \mu \in K$.
Let $A, B \subseteq X$.
Then:
:$T \sqbrk {\lambda A + \mu B} = \lambda T \sqbrk A + \mu T \sqbrk B$
== Proof ==
We have: 
Category:Linear Combination of Subsets of Vector Spaces
Image of 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 $\Img \phi$ denote the image set of $\phi$.
Then $\Img \phi$ is a submodule of $H$.
== Proof ==
By Module is Submodule of Itself, $\struct {G, +_G, \circ_G}_R$ is a submodule of $\struct {G, +_G, \circ_G}_R$.
The result follows from Image of Submodule under Linear Transformation is Submodule.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.2$
Category:Linear Transformations
Image of Mapping/Examples/Image of x^2-4x+5

== Example of Image of Element under Mapping ==
Let $f: \R \to \R$ be the mapping defined as:
:$\forall x \in \R: \map f x = x^2 - 4 x + 5$
The image of $f$ is the unbounded closed interval:
:$\Img f = \hointr 1 \to$
and so $f$ is not a surjection.
=== Graphical Representation of $\map f x = x^2 - 4 x + 5$ ===
== Proof ==
By differentiating $x^2 - 4 x + 5$ twice $x$:
:$f' = 2 x - 4$
:$f' = 2 x - 4$
Equating $f'$ to $0$, a stationary point is found at $x = 2$.
Inspecting the sign of $f$, it is noted that $f'$ is increasing everywhere.
Hence the stationary point at $x = 2$ is a minimum of $\Img f$.
This is the only stationary point, so it can be stated that the minimum of $f$ occurs at $x = 2$.
We have that:
:$f \paren 2 = 2^2 - 4 \times 2 + 5 = 4 - 8 + 5 = 1$
As $f$ is strictly increasing on $x > 2$ and strictly decreasing on $x < 2$, it is seen that $f$ is unbounded above.
Thus:
:$\Img f = \hointr 1 \to$
== Also see ==
* Bijective Restrictions of $\map f x = x^2 - 4 x + 5$
== Sources ==
* : Chapter $4$: Mappings: Exercise $12 \ \text{(ii)}$
Category:Examples of Images under Mappings
Image of Mapping/Examples/Image of x^4-1

== Example of Image of Element under Mapping ==
Let $f: \R \to \R$ be the mapping defined as:
:$\forall x \in \R: \map f x = x^4 - 1$
The image of $f$ is the unbounded closed interval:
:$\Img f = \hointr {-1} \to$
and so $f$ is not a surjection.
== Proof ==
Trivially, by differentiating $x^4 - 1$ $x$:
:$f' = 4 x^3$
and equating $f'$ to $0$, the minimum of $\Img f$ is seen to occur at $f \paren 0 = -1$.
Thus it can be seen that the minimum of $\Img f$ is $-1$.
As $f$ is strictly increasing on $x > 0$ and strictly decreasing on $x < 0$, it is seen that $f$ is unbounded above.
Thus:
:$\Img f = \hointr {-1} \to$
As $\Img f \subset \R$ and $\Img f \ne \R$, it follows by definition that $f$ is not surjective.
== Sources ==
* : Chapter $4$: Mappings: Exercise $3 \ \text {(iii)}$
Category:Examples of Images under Mappings
Image of Mapping from Finite Set is Finite

== Theorem ==
Let $X, Y$ be sets.
Let $f: X \to Y$ be a mapping.
Let $X$ be a finite set.
Then $f \sqbrk X$ is a finite set.
== Proof ==
By definition of surjection:
:$f: X \to f \sqbrk X$ is a surjection.
The case when $X \ne \O$:
By Surjection iff Cardinal Inequality:
:$\card {f \sqbrk X} \le \card X$
Thus by Set of Cardinality not Greater than Cardinality of Finite Set is Finite:
:$f \sqbrk X$ is finite.
The case when $X = \O$:
By a corollary to Image of Empty Set is Empty Set:
:$f \sqbrk X = \O$
Thus the result holds.
== Sources ==
* 
Category:Images
Category:Finite Sets
Image of Mapping from Group Element to Inner Automorphism is Inner Automorphism Group

== Theorem ==
Let $G$ be a group.
Let $\kappa: G \to \Aut G$ be the mapping from $G$ to the automorphism group of $G$ defined as:
:$\forall x \in G: \map \kappa x := \kappa_x$
where $\kappa_x$ is the inner automorphism on $x$:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
Then $\Img \kappa$ is the inner automorphism group of $G$.
== Proof ==
Let $\Inn G$ denote the inner automorphism group of $G$.
For all $x \in G$, $\map \kappa x = \kappa_x \in \Inn G$.
Hence $\Img \kappa \subseteq \Inn G$. 
Let $\phi \in \Inn G$. Then:
:$\exists y \in G: \forall g \in G: \map \phi g = y g y^{-1}$
Then $\map \kappa y = \phi$.
Hence $\Inn G \subseteq \Img \kappa$. 
Therefore $\Img \kappa = \Inn G$.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $10$
Category:Inner Automorphisms
Category:Group Homomorphisms
Image of Mapping of Infima is Generator Set of Filter

== Theorem ==
Let $\left({S, \wedge, \preceq}\right)$ be a meet semilattice.
Let $f, g:\N \to S$ be mappings such that
:$\forall n \in \N: g\left({n}\right) = \inf \left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$
Let $F$ be a filter such that
:$f\left[{\N}\right]$ is generator set of $F$,
where $f\left[{\N}\right]$ denotes the image of $f$.
Then $g\left[{\N}\right]$ is generator set of $F$.
== Proof ==
By Set is Coarser than Image of Mapping of Infima:
:$f\left[{\N}\right]$ is coarser than $g\left[{\N}\right]$.
By definition of generator set of filter:
:$F = \left({\operatorname{fininfs}\left({f\left[{\N}\right]}\right)}\right)^\succeq$
where
:$\operatorname{fininfs}\left({f\left[{\N}\right]}\right)$ denotes the finite infima set of $f\left[{\N}\right]$,
:for subset $A$ of $S$: $A^\succeq$ denotes upper closure of $A$.
By Finite Infima Set and Upper Closure is Smallest Filter:
:$f\left[{\N}\right] \subseteq F$
We will prove that
:$g\left[{\N}\right] \subseteq F$
Let $x \in g\left[{\N}\right]$.
By definition of image of mapping:
:$\exists n \in \N: x = g\left({n}\right)$
By definition of $g$:
:$x = \inf \left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$
By definition:
:$\left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$ is a finite set.
By definition of subset:
:$\left\{ {f\left({m}\right): m \in \N \land m \le n}\right\} \subseteq F$
By definition of reflexivity:
:$f\left({n}\right) \in \left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$
Thus by Filtered in Meet Semilattice with Finite Infima:
:$x \in F$
By Subset of Set is Coarser than Set
:$g\left[{\N}\right]$ is coarser than $F$.
Thus by Coarser Between Generator Set and Filter is Generator Set of Filter:
:$g\left[{\N}\right]$ is generator set of $F$.
== Sources ==
* 
* 
Category:Join and Meet Semilattices
Image of Mapping of Intersections is Smallest Basis

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological space.
Let $f: X \to \tau$ be a mapping such that:
:$\forall x \in X: \paren {x \in \map f x \land \forall U \in \tau: x \in U \implies \map f x \subseteq U}$.
Then the image $\Img f$ is subset of every basis of $T$.
== Proof ==
Let $\BB$ be a basis.
Let $V \in \Img f$.
Then by definition of image there exists a point $b \in X$ such that:
:$V = \map f b$
Then $V$ is open because $\Img f \subseteq \tau$.
By assumption of mapping $f$:
:$b \in V$
Then by definition of basis there exists a subset $U \in \BB$ such that:
:$b \in U \subseteq V$
By definition of basis:
:$\BB \subseteq \tau$
Then by definition of subset:
:$U \in \tau$
Then by assumption of mapping $f$:
:$\map f b \subseteq U$
Thus by definition of set equality:
:$V = U \in \BB$
== Sources ==
* 
* 
Category:Topology
Image of Open Set under Continuous Mapping in Metric Space may not be Open

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a $\tuple {d_1, d_2}$-continuous mapping from $A_1$ to $A_2$.
Let $U \subseteq A_1$ be an open set of $M_1$.
Then it is not necessarily the case that $f \sqbrk U$ is an open set of $M_2$.
== Proof ==
Consider the constant mapping $f_0: \R^2 \to \R$ defined as:
:$\forall \tuple {x, y} \in \R^2: \map f {x, y} = 0$
Then by Constant Mapping is Continuous $f$ is a continuous mapping.
But consider any open set $U \subseteq A_1$ of $M_1$ such that $U \ne \O$.
Then $f \sqbrk U = \set 0 = \closedint 0 0$ which is a closed interval of $\R$.
The result follows from Closed Real Interval is not Open Set.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Warning $2.3.14$
Category:Open Sets (Metric Spaces)
Category:Continuity
Image of Operator Generated by Closure System is Set of Closure System

== Theorem ==
Let $L = \struct {X, \vee, \wedge, \preceq}$ be a complete lattice.
Let $S = \struct {T, \precsim}$ be a closure system of $L$.
Then $\map {\operatorname {operator} } S \sqbrk X = T$
where $\map {\operatorname {operator} } S$ denotes the operator generated by $S$.
== Proof ==
Define $f = \map {\operatorname {operator} } S$.
=== $\subseteq$ ===
Let $x \in f \sqbrk X$.
By definition of image of mapping:
:$\exists y \in X: x = \map f y$
Define $Y = y^\succeq \cap T$
By definition of complete lattice:
:$Y$ admits an infimum in $L$.
By Intersection is Subset:
:$Y \subseteq T$
By definition of closure system:
:$S$ inherits infima.
By definition of infima inheriting:
:$Y$ admits an infimum in $S$ and $\inf_S Y = \inf_L Y$
Thus by definition of operator generated by $S$:
:$x \in T$
=== $\supseteq$ ===
Let $x \in T$.
By definition of ordered subset:
:$T \subseteq X$
By definition of subset:
:$x \in X$
Define $Y = x^{\succeq_L} \cap T$
By definitions of upper closure of element and reflexivity:
:$x \in x^{\succeq_L}$
By definition of intersection:
:$x \in Y$
By definitions of infimum and lower bound:
:$\inf_L Y \preceq x$
By Intersection is Subset:
:$Y \subseteq x^{\succeq_L}$
By Infimum of Subset:
:$\inf_L x^{\succeq_L} \preceq \inf_L Y$
By Infimum of Upper Closure of Element:
:$x \preceq \inf_L Y$
By definition of antisymmetry:
:$x = \inf_L Y$
By definition of operator generated by $S$:
:$x = \map f x$
Thus by definition of image of mapping:
:$x \in f \sqbrk X$
Hence by definition of set equality:
:$f \sqbrk X = T$
== Sources ==
* 
* 
Category:Closure Operators
Image of P-adic Norm

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.
Then the image of $\norm {\,\cdot\,}_p$ is:
:$\Img {\norm {\,\cdot\,}_p} = \set {p^n : n \in \Z} \cup \set 0$
== Proof ==
This follows immediately from:
:
:
Category:P-adic Number Theory
Image of Path is Path-Connected Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $I \subset \R$ be the closed real interval $\closedint a b$.
Let $\gamma: I \to S$ be a path.
Then:
:$\map \gamma I$ is a path-connected set of $T$.
== Proof ==
From Path-Connected iff Path-Connected to Point, $\map \gamma I$ is a path-connected set every point of $\map \gamma I$ is path-connected to a common point.
It is shown that every point of $\map \gamma I$ is path-connected to $\map \gamma a$.
Let $H = \map \gamma I$.
Let $x = \map \gamma a$.
From Point is Path-Connected to Itself, $x$ is path-connected to itself.
Let $y$ be any element of $H \setminus \set x$.
By assumption:
:$\exists t \in \hointl a b : \map \gamma t = y$
From Continuity of Composite with Inclusion, $\mathbin \gamma {\restriction_{\closedint a t} } : \closedint a t \to H$ is a continuous mapping.
It follows that $\mathbin \gamma {\restriction_{\closedint a t} }$ is a path joining $x$ to $y$.
The result follows.
Category:Path-Connected Spaces
Image of Point under Central Dilatation Mapping with Fixed Point at Origin

== Theorem ==
Let $f$ be a central dilatation mapping whose center of enlargement $C$ is located at the origin of a coordinate system.
Let the scale factor of $f$ be $k$.
Let the position vector of a point $P$ be $\mathbf r$.
Then the image of $P$ under $f$ has position vector $k \mathbf r$.
=== Corollary ===
== Proof ==
== Sources ==
* 
* 
Category:Image of Point under Central Dilatation Mapping with Fixed Point at Origin
Category:Central Dilatation Mappings
Image of Point under Central Dilatation Mapping with Fixed Point at Origin/Corollary

== Theorem ==
Let $f$ be a central dilatation mapping whose center of enlargement $C$ is located at the origin of a Cartesian plane.
Let the scale factor of $f$ be $k$.
Let $P$ be the point $\tuple {x, y}$.
Then the image of $P$ under $f$ is $\tuple {k x, k y}$.
== Proof ==
== Sources ==
* 
* 
Category:Image of Point under Central Dilatation Mapping with Fixed Point at Origin
Image of Point under Negative Scale Factor

== Theorem ==
Let $f$ be a central dilatation mapping with center of enlargement $C$.
Let the scale factor of $f$ be negative.
Let the image of a point $P$ be $P'$.
Then $PCP'$ is a straight line such that $C$ is between $P$ and $P'$.
== Proof ==
== Sources ==
* 
* 
Category:Central Dilatation Mappings
Image of Point under Neighborhood of Diagonal is Neighborhood of Point

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $V$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:$\forall x \in X : \map V x$ is a neighborhood of $x$ in $T$
== Proof ==
Let $x \in X$.
By definition of diagonal:
:$\tuple{x, x} \in \Delta_X$
From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$V$ is a neighborhood of $\tuple{x, x}$
By definition of the product topology:
:$\BB = \set {U_1 \times U_2: U_1, U_2 \in \tau}$ is a basis for $\tau_{X \times X}$
From Characterization of Neighborhood by Basis:
:$\exists U_1, U_2 \in \tau : \tuple{x, x} \in U_1 \times U_2$ and $U_1 \times U_2 \subseteq V$
By definition of Cartesian product:
:$x \in U_1, x \in U_2$
We have:
Hence $\map V x$ is a neighborhood of $x$ in $T$ by definition.
The result follows.
Category:Neighborhoods
Category:Product Spaces
Category:Relations
Image of Point under Open Neighborhood of Diagonal is Open Neighborhood of Point

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $U$ be an open neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:$\forall x \in X : \map U x$ is an open neighborhood of $x$ in $T$
== Proof ==
Let $x \in X$.
By definition of open neighborhood:
:$U$ is a neighborhood of $\Delta_X$
From Image of Point under Neighborhood of Diagonal is Neighborhood of Point:
:$\map U x$ is a neighborhood of $x$ in $T$
It remains to show that $\map U x$ is open in $T$, that is, $\map U x \in \tau$.
Let:
:$\WW = \set{W \in \tau : \exists V \in \tau : x \in V, V \times W \subseteq U}$.
Let $y \in \map U x$.
By definition of image:
:$\tuple {x, y} \in U$
By definition of product topology:
:$\BB = \set{V \times W : V, W \in \tau}$ is a basis for $\tau_{X \times X}$
By definition of basis:
:$\exists V, W \in \tau : \tuple{x, y} \in V \times W \subseteq U$
By definition of $\WW$:
:$W \in \WW$
Since $y$ was arbitrary:
:$\forall y \in \map U x : \exists W \in \WW : y \in W$
Let $W \in \WW$.
By definition of $\WW$:
:$\exists V \in \tau : x \in V, V \times W \subseteq U$
From Cartesian Product of Subsets:
:$\set x \times W \subseteq V \times W$
From Subset Relation is Transitive:
:$\set x \times W \subseteq U$
By definition of image:
:$W \subseteq \map U x$
Since $W$ was arbitrary:
:$\forall W \in \WW : W \subseteq \map U x$
From Characterization of Set Equals Union of Sets:
:$\map U x = \bigcup \WW$
By :
:$\map U x \in \tau$
Since $x$ was arbitrary:
:$\forall x \in X : \map U x$ is an open neighborhood of $x$ in $T$
Category:Open Sets
Category:Product Spaces
Category:Relations
Image of Pointwise Scalar Multiplication of Subset of Scalars with Subset of Vectors under 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 $S \subseteq K$ and $D \subseteq X$ be non-empty sets.
Then:
:$T \sqbrk {S D} = S T \sqbrk D$
where:
:$S D = \set {\lambda x : \lambda \in S, \, x \in D}$
== Proof ==
We have: 
Category:Vector Spaces
Image of Positive Element of C*-Algebra under *-Algebra Homomorphism is Positive

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,}_A}$ and $\struct {B, \square, \norm {\, \cdot \,}_B}$ be $\text C^\ast$-algebras.
Let $\phi : A \to B$ be a $\ast$-algebra homomorphism.
Let $a \in A$ be positive.
Then $\map \phi a$ is positive.
== Proof ==
We first show that $\map \phi a \in B$ is Hermitian.
Since $\phi$ is a $\ast$-algebra isomorphism, we have:
:$\map \phi a^\square = \map \phi {a^\ast} = \map \phi a$
Hence $\map \phi a$ is Hermitian.
From Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism: Corollary, we have:
:$\map {\sigma_B} {\map \phi a} \subseteq \map {\sigma_A} a \cup \set 0 \subseteq \hointr 0 \infty$
Hence $\map \phi a$ is positive.
Category:Positive Elements of C*-Algebras
Image of Preimage of Ideal under Ring Epimorphism

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $S_2$ be an ideal of $R_2$.
Then:
:$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$
== Proof ==
As an ideal is a subring, the result Image of Preimage of Subring under Ring Epimorphism applies directly.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 3^\circ$
Category:Ring Epimorphisms
Category:Subrings
Image of Preimage of Subring under Ring Epimorphism

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $S_2$ be a subring of $R_2$.
Then:
:$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$
== Proof ==
As $\phi$ is an epimorphism, it is a surjection, and so:
:$\Img \phi = R_2$
So:
:$S_2 \subseteq \Img {R_1}$
The result then follows from Image of Preimage under Mapping.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 3^\circ$
Category:Ring Epimorphisms
Category:Subrings
Image of Preimage of Subset under Surjection equals Subset

== Theorem ==
Let $f: S \to T$ be a surjection.
Then:
:$\forall B \subseteq T: B = \paren {f \circ f^{-1} } \sqbrk B$
where:
:$f \sqbrk B$ denotes the image of $B$ under $f$
:$f^{-1}$ denotes the inverse of $f$
:$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.
== Proof ==
Let $g$ be a surjection.
Let $B \subseteq T$.
Let $b \in B$.
Then:
From Subset of Codomain is Superset of Image of Preimage, we already have that:
:$\paren {f \circ f^{-1} } \sqbrk B \subseteq B$
So:
:$B \subseteq \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
and by definition of set equality:
:$B = \paren {f \circ f^{-1} } \sqbrk B$
== Also see ==
* Subset equals Image of Preimage implies Surjection
* Subset equals Image of Preimage iff Mapping is Surjection
* Subset of Domain is Subset of Preimage of Image
== 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 $1 \ \text{(d})$
* : $1$: Set Theory and Logic: $\S 2$: Functions
Category:Surjections
Category:Preimages under Mappings
Category:Composite Mappings