nyuuzyou/wikis huggingface dataset:
Image of Composite Mapping

== Theorem ==
Let $f: S \to T$ and $g: R \to S$ be mappings.
Then:
:$\Img {f \circ g} = f \sqbrk {\Img g}$
where:
: $f \circ g$ is the composition of $g$ and $f$
: $\operatorname{Img}$ denotes image
: $f \sqbrk \cdot$ denotes taking image of a subset under $f$.
=== Corollary ===
== Proof ==
By definition of image:
:$\Img {f \circ g} = \set {t \in T: \exists r \in R: \map {\paren {f \circ g} } r = t}$
and by definition of the image of a subset:
:$f \sqbrk {\Img g} = \set {t \in T: \exists s \in \Img g: \map f s = t}$
which, expanding what it means that $s \in \Img g$, equals:
:$f \sqbrk {\Img g} = \set {t \in T: \exists s \in S: \exists r \in R: \map g r = s \land \map f s = t}$
Now substituting $\map g r = s$ in $\map f s = t$, we obtain:
:$f \sqbrk {\Img g} = \set {t \in T: \exists s \in S: \exists r \in R: \map f {\map g r} = t}$
which is seen to equal the expression for $\Img {f \circ g}$ as soon as $S$ is non-empty.
The remaining case to be checked is thus if $S = \O$.
From Null Relation is Mapping iff Domain is Empty Set, also $R = \O$, so that $f$ and $g$ are empty mappings.
By Image of Empty Set is Empty Set, we conclude:
:$\Img {f \circ g} = \Img g = \O$
and also:
:$f \sqbrk {\Img g} = \O$
which together yield the desired equality.
Hence the result.
== Also see ==
* Image of Composite Relation
Category:Composite Mappings
Image of Composite Mapping/Corollary

== Corollary to Image of Composite Mapping ==
Let $f: S \to T$ and $g: R \to S$ be mappings.
Then:
:$\Img {f \circ g} \subseteq \Img f$
where:
:$f \circ g$ denotes composition of $g$ and $f$
:$\Img f$ denotes image of $f$.
== Proof ==
From Image of Composite Mapping, it holds that:
:$\Img {f \circ g} = f \sqbrk {\Img g}$
where $f \sqbrk {\, \cdot \,}$ denotes image of subset.
By definition of composite mapping:
:$\Img g \subseteq \Dom f$
where $\Dom f$ denotes the domain of $f$.
From Image of Subset under Mapping is Subset of Image:
:$\Img {f \circ g} \subseteq \Img f$
== Sources ==
* : $\S 1.14$: Composition of Functions
Category:Composite Mappings
Image of Composite Relation

== Theorem ==
Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be relations.
Let $\RR_2 \circ \RR_1 \subseteq S_1 \times T_2$ be the composition of $\RR_1$ and $\RR_2$.
Then the image of $\RR_2 \circ \RR_1$ is given by:
:$\Img {\RR_2 \circ \RR_1} = \Img {\Img {\RR_1} \cap \Preimg {\RR_2} }$
== Proof ==
We have by definition of composition of $\RR_1$ and $\RR_2$:
:$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
Let $\left({x, z}\right) \in \RR_2 \circ \RR_1$.
By definition of image:
:$z \in \Img {\RR_2 \circ \RR_1}$
By definition of composition of $\RR_1$ and $\RR_2$:
:$z \in \Img {\RR_2}$
But also:
:$z \in \Img {\Img {\RR_1} }$
We also have that $\Img {\RR_2} = \Img {\Preimg {\RR_2} }$
That is:
$z \in \Img {\Img {\RR_1} } \cap \Img {\Preimg {\RR_2} }$
Category:Composite Relations
Image of Computable Real Number over Sequentially Computable Real Function is Computable

== Theorem ==
Let $f : \R \to \R$ be a sequentially computable real function.
Let $a \in \R$ be a computable real number.
Then, $\map f a$ is a computable real number.
== Proof ==
Let $\sequence {x_n}$ be defined as:
:$x_n = a$
By Constant Sequence of Computable Real Number is Computable:
:$\sequence {x_n}$ is a computable real sequence.
As $f$ is sequentially computable:
:$\sequence {\map f {x_n}}$ is computable.
By Term of Computable Real Sequence is Computable:
:$\map f {x_0}$ is computable.
But, $x_0 = a$ by definition.
Thus:
:$\map f a$ is computable.
Category:Computability Theory
Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded

== Theorem ==
Let $\struct {X, \tau}$ be a topological vector space. 
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence with $x_n \to x$. 
Let: 
:$E = \set {x_n : n \in \N}$
Then $E$ is von Neumann-bounded. 
== Proof 1 ==
== Proof 2 ==
Category:von Neumann-Bounded Subsets of Topological Vector Spaces
Category:Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded
Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded/Proof 1

== Theorem ==
== Proof ==
From Convergent Sequence in Topological Vector Space is Cauchy, $\sequence {x_n}_{n \in \N}$ is a Cauchy sequence.
From Image of Cauchy Sequence in Topological Vector Space is von Neumann-Bounded, $E$ is von Neumann-bounded.
Category:Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded
Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded/Proof 2

== Theorem ==
== Proof ==
From Union of Image of Convergent Sequence and Limit in Topological Space is Compact, $E \cup \set x$ is compact.
From Compact Subspace of Topological Vector Space is von Neumann-Bounded, $E \cup \set x$ is von Neumann bounded.
From Subset of von Neumann-Bounded Set is von Neumann-Bounded, $E$ is von Neumann bounded.
Category:Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded
Image of Convergent Sequence in Topological Vector Space under Bounded Linear Transformation is von Neumann-Bounded

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be topological vector spaces over $\GF$. 
Let $T : X \to Y$ be a bounded linear transformation.
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $X$. 
Then $\set {T x_n : n \in \N}$ is a von Neumann-bounded subset of $Y$. 
== Proof ==
Let:
:$E = \set {x_n : n \in \N}$
From Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded, $E$ is von Neumann-bounded.
So, from the definition of a bounded linear transformation:
:$T \sqbrk E = \set {T x_n : n \in \N}$ is von Neumann-bounded.
== Sources ==
* : $1.32$: Theorem
Category:Bounded Linear Transformations (Topological Vector Spaces)
Image of Convex Set under Linear Transformation is Convex

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\Bbb F$.
Let $C \subseteq X$ be convex.
Let $T : X \to Y$ be a linear transformation.
Then $\map T C \subseteq Y$ is convex.
== Proof ==
Let $y_1, y_2 \in \map T C$ and $\lambda \in \closedint 0 1$. 
Then, there exists $x_1, x_2 \in C$ such that: 
:$y_1 = T x_1$
and:
:$y_2 = T x_2$ 
Then, we have: 
:$\lambda y_1 + \paren {1 - \lambda} y_2 = \lambda T x_1 + \paren {1 - \lambda} T x_2 = \map T {\lambda x_1 + \paren {1 - \lambda} x_2}$
Since $C$ is convex, we have: 
:$\lambda x_1 + \paren {1 - \lambda} x_2 \in C$
So, we have: 
:$\map T {\lambda x_1 + \paren {1 - \lambda} x_2} \in \map T C$
That is: 
:$\lambda y_1 + \paren {1 - \lambda} y_2 \in \map T C$
So $\map T C$ is convex.
Category:Convex Sets (Vector Spaces)
Category:Linear Transformations
Image of Countable Set under Mapping is Countable

== Definition ==
Let $S$ be a countable set.
Let $T$ be a set.
Let $f: S \to T$ be a mapping.
Then the image of $f$ is countable.
== Proof ==
Let $A$ be the image of $f$.
Let $g: S \to A$ be the restriction of $f$ to the cartesian product $S \times A$.
By Surjection by Restriction of Codomain, $g$ is a surjection.
By Surjection from Natural Numbers iff Countable, there exists a surjection $\phi: \N \to S$.
Since the composition of surjections is a surjection, $g \circ \phi: \N \to A$ is a surjection.
Hence, by Surjection from Natural Numbers iff Countable, it follows that $A$ is countable.
Category:Countable Sets
Category:Images
Image of Dilation of Set under Linear Transformation is Dilation of Image

== Theorem ==
Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $\lambda \in K$. 
Then: 
:$\map T {\lambda E} = \lambda \map T E$
where $\lambda E$ denotes the dilation of $E$ by $\lambda$.
== Proof ==
We have: 
:$y \in \map T {\lambda E}$
:
:$y = \map T {\lambda x}$ for some $x \in E$.
From the linearity of $T$, this is equivalent to:
:$y = \lambda T x$
This is equivalent to:
:$y \in \lambda \map T E$
So by the definition of set equality we have:
:$\map T {\lambda E} = \lambda \map T E$
Category:Linear Transformations
Category:Dilation of Subsets of Vector Spaces
Image of Directed Subset under Increasing Mapping is Directed

== Theorem ==
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be ordered sets.
Let $f: S \to T$ be an increasing mapping.
Let $D$ be a directed subset of $S$.
Then
:$f^\to \left({D}\right)$ is also a directed subset of $T$
where
:$f^\to \left({D}\right)$ denotes the image of $D$ under $f$.
== Proof ==
Let $x, y \in f^\to\left({D}\right)$.
By definition of image of set:
:$\exists a \in D: x = f \left({a}\right)$
and
:$\exists b \in D: y = f \left({b}\right)$
By definition of directed subset:
:$\exists c \in D: a \preceq c \land b \preceq c$
By definition of image of set:
:$f\left({c}\right) \in f^\to\left({D}\right)$
By definition of increasing mapping:
:$x \precsim f\left({c}\right)$ and $y \precsim f\left({c}\right)$
Thus by definition
:$f^\to\left({D}\right)$ is a directed subset of $T$.
== Sources ==
* 
Category:Order Theory
Image of Domain of Mapping is Image Set

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
The image of $S$ is the image set of $f$:
:$f \sqbrk S = \Img f$
== Proof ==
By definition, a mapping is a relation.
Thus Image of Domain of Relation is Image Set applies.
== Sources ==
* : $\S 8$: Functions
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms
* : Chapter $\text{I}$: Sets and Functions: Functions
* : $\S 1.10$: Functions: Remark $10.8 \ \text{(b)}$
Category:Images
Image of Domain of Relation is Image Set

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
The image of the domain of $\RR$ is the image set of $\RR$:
:$\RR \sqbrk {\Dom \RR} = \Img \RR$
where $\Img \RR$ is the image of $\RR$.
== Proof ==
Let $y \in \RR \sqbrk {\Dom \RR}$.
Let $y \in \Img \RR$.
The result follows by definition of set equality.
Category:Relation Theory
Image of Doubleton under Mapping

== Theorem ==
Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall x, y \in S: f \sqbrk {\set {x, y} } = \set {\map f x, \map f y}$
== Proof ==
Let $x, y \in S$.
Thus
== Sources ==
* 
Category:Images
Category:Doubletons
Image of Element is Subset

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $A \subseteq S$.
Then:
:$s \in A \implies \map \RR s \subseteq \RR \sqbrk A$
== Proof ==
From Image of Singleton under Relation:
:$\map \RR s = \RR \sqbrk {\set s}$
From Singleton of Element is Subset:
:$s \in A \implies \set s \subseteq A$
The result follows from Image of Subset is Subset of Image.
Category:Relation Theory
Image of Element under Cartesian Product of Subsets

== Theorem ==
Let $S$ and $T$ be sets.
Let $A \subseteq S$ and $B \subseteq T$.
Let $\RR$ be the relation defined by the Cartesian product $A \times B$.
Then:
:$\forall x \in A: \map \RR x = B$
== Proof ==
We have:
Category:Cartesian Product
Category:Relations
Image of Element 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 $x \in S_1$.
Then:
:$\RR_2 \sqbrk {\map {\RR_1} x \cap S_2} = \map {\paren{\RR_2 \circ \RR_1}} x$
== Proof ==
We have:
== Also see ==
* Image of Element under Composite Relation with Common Codomain and Domain
* Image of Subset under Composite Relation
* Image of Subset under Composite Relation with Common Codomain and Domain
Category:Composite Relations
Image of Element 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 $x \in S$.
Then:
:$\RR_2 \sqbrk {\map {\RR_1} x} = \map {\paren{\RR_2 \circ \RR_1}} x$
== Proof ==
We have:
== Also see ==
* Image of Subset under Composite Relation with Common Codomain and Domain
* Image of Element under Composite Relation
* Image of Subset under Composite Relation
Category:Composite Relations
Image of Element under Inverse Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping such that its inverse $f^{-1}: T \to S$ is also a mapping.
Then:
:$\forall x \in S, y \in T: \map f x = y \iff \map {f^{-1} } y = x$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
=== Sufficient Condition ===
Let $f: S \to T$ be a mapping.
From the definition of inverse mapping:
:$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$
Let $y = \map f x$.
From the definition of the preimage of an element:
:$\map {f^{-1} } y = \set {s \in S: \tuple {y, x} \in f}$
Thus:
:$x \in \map {f^{-1} } y$
However, $f^{-1}$ is a mapping.
Therefore, by definition:
:$\forall y \in T: \tuple {y, x_1} \in f^{-1} \land \tuple {y, x_2} \in f^{-1} \implies x_1 = x_2$
Thus:
:$\forall s \in \map {f^{-1} } y: s = x$
Thus:
:$\map {f^{-1} } y = \set x$
That is:
:$x = \map {f^{-1} } y$
=== Necessary Condition ===
Let $\map {f^{-1} } y = x$.
By definition of inverse mapping:
:$\map f x = y$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.2$
* : $\S 22.3$: Injections; bijections; inverse of a bijection
Category:Image of Element under Inverse Mapping
Category:Inverse Mappings
Image of Element under Inverse Mapping/Corollary 1

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping such that its inverse $f^{-1}: T \to S$ is also a mapping.
Then:
:$\forall x \in S: \map {f^{-1} } {\map f x} = x$
== Proof ==
== Sources ==
* 
* 
Category:Image of Element under Inverse Mapping
Image of Element under Inverse Mapping/Corollary 2

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping such that its inverse $f^{-1}: T \to S$ is also a mapping.
Then:
:$\forall y \in T: \map f {\map {f^{-1} } y} = y$
== Proof ==
== Sources ==
* 
* 
Category:Image of Element under Inverse Mapping
Image of Element under Mapping/Examples/Image of 2 under 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 $2$ is:
:$\map f 2 = 15$
== Proof ==
:$2^4 - 1 = 16 - 1 = 15$
== Sources ==
* : Chapter $4$: Mappings: Exercise $3 \ \text {(i)}$
Category:Examples of Images under Mappings
Image of Element under Mapping/Examples/Images of Various Numbers under x^2+2x+1 in Limited Range

== Examples of Images of Elements under Mapping ==
Let $f: \closedint 0 1 \to \R$ be the mapping defined as:
:$\forall x \in \closedint 0 1: \map f x = x^2 + 2 x + 1$
where $\closedint 0 1$ denotes the closed real interval from $0$ to $1$.
The images of various real numbers under $f$ are:
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text B$: The Meaning of the Term Function of One Independent Variable: Example $2.5$
Category:Examples of Images under Mappings
Image of Empty Set is Empty Set/Corollary 1

== Corollary of Image of Empty Set is Empty Set ==
Let $f: S \to T$ be a mapping.
The image of the empty set is the empty set:
:$f \sqbrk \O = \O$
== Proof ==
By definition, a mapping is a relation.
Thus Image of Empty Set is Empty Set applies.
== Sources ==
* : $\S 1.10$: Functions: Remark $10.8 \ \text{(a)}$
Category:Images
Category:Empty Set
Image of Empty Set is Empty Set/Corollary 2

== Corollary of Image of Empty Set is Empty Set ==
Let $S = \O$ and $T \ne \O$.
There is no surjection $f: S \to T$.
== Proof ==
Let $f: S \to T$.
By Image of Empty Set is Empty Set: Corollary 1:
:$f \sqbrk S = \O$
Thus:
:$f \sqbrk S \ne T$
and $f$ is not a surjection.
Category:Surjections
Category:Empty Set
Image of Evaluation Linear Transformation on Banach Space is Closed

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ be a Banach space.
Let $\struct {X^{\ast \ast}, \norm \cdot_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm \cdot_X}$.
Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then: 
:$\map J X$ is closed in $X^{\ast \ast}$.
== Theorem ==
Let $L$ be the limit of a convergent sequence in $\map J X$.
Let $\sequence {j_n}_{n \mathop \in \N}$ be a sequence in $\map J X$ such that: 
:$\sequence {j_n}_{n \mathop \in \N}$ converges to $L$.
Note that for each $n \in \N$ there exists $x_n \in X$ such that: 
:$j_n = J x_n$
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have: 
:$J$ is a linear isometry.
Specifically:
:$J$ is a bounded linear transformation.
From Continuity of Linear Transformations, we have: 
:$J$ is continuous.
We show that $\sequence {x_n}_{n \mathop \in \N}$ converges to some $x \in X$.
We will then have:
:$L = J x \in \map J X$ 
from continuity.
Note that: 
Let $\epsilon > 0$. 
Since: 
:$\sequence {j_n}_{n \mathop \in \N}$ is convergent
we have, from Convergent Sequence in Normed Vector Space is Cauchy Sequence:
:there exists $N \in \N$ such that for all $n, m \ge N$, we have $\norm {j_n - j_m}_X < \epsilon$.
So, for $n, m \ge N$, we have: 
:$\norm {x_n - x_m}_X < \epsilon$
Since $\epsilon$ was arbitrary, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Since $X$ is a Banach space, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges to a limit $x \in X$.
Since $J$ is continuous, we obtain:
Since: 
:$J x \in \map J X$
we have: 
:$L \in \map J X$
Since $L$ was an arbitrary limit in $\map J X$, we have: 
:$\map J X$ contains each limit of a convergent sequence in $\map J X$
So, from Subset of Metric Space contains Limits of Sequences iff Closed:
:$\map J X$ is closed in $X^{\ast \ast}$.
== Sources ==
* $26.1$: The Second Dual
Category:Evaluation Linear Transformations (Normed Vector Spaces)
Category:Linear Transformations on Banach Spaces
Image of Evaluation Linear Transformation on Banach Space is Closed Linear Subspace of Second Dual

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ be a Banach space.
Let $\struct {X^{\ast \ast}, \norm \cdot_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm \cdot_X}$.
Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then $\map J X$ is a closed linear subspace of $X^{\ast \ast}$. 
== Proof ==
From Image of Vector Subspace under Linear Transformation is Vector Subspace:
:$\map J X$ is a vector subspace of $X^{\ast \ast}$. 
From Image of Evaluation Linear Transformation on Banach Space is Closed:
:$\map J X$ is closed in $X^{\ast \ast}$.
Category:Evaluation Linear Transformations (Normed Vector Spaces)
Image of Fredholm Operator of Banach Spaces is Closed

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\tuple {X, \norm \cdot_X}, \tuple {Y, \norm \cdot_Y}$ be Banach spaces over $\Bbb F$.
Let $T: X \to Y$ be a Fredholm operator.
Let $\Img T$ be the image of $T$ .
Then $\Img T$ is closed.
== Proof ==
By Kernel of Bounded Linear Transformation is Closed Linear Subspace, $\map \ker T \subseteq X$ is closed.
Thus, by Characterization of Complete Normed Quotient Vector Spaces, $X / \map \ker T$ is a Banach space.
Define the injective linear transformation $\tilde T : X / \map \ker T \to Y$ by:
:$\map {\tilde T} {\eqclass x {\map \ker T} } := \map T x$
Then $\Img {\tilde T} = \Img T$.
Replacing $T$ by $\tilde T$ if necessary, we may assume that $T$ is injective.
Since $Y / \Img T$ is finite-dimensional, it admits a finite basis:
:$\set { \eqclass {y_1} {\Img T}, \ldots, \eqclass {y_m} {\Img T}}$
Consider the direct sum:
:$X \oplus {\Bbb F^m}$
Define the surjective linear transformation $\hat T : X \oplus {\Bbb F^m} \to Y$ by:
:$\map {\hat T} {x, k_1, \ldots, k_m } := \map T x + k_1 y_1 + \cdots + k_m y_m$
Observe:
where:
:$\norm T$ denotes the operator norm of $T$
:$L := \sqrt {\norm {y_1}_Y^2 + \cdots + \norm {y_m}_Y^2}$
Hence $\hat T$ is bounded.
Furthermore, $\hat T$ is injective.
Indeed, let
:$\map {\hat T} {x, k_1, \ldots, k_m } = 0$
That is:
:$k_1 y_1 + \cdots + k_m y_m = - \map T x$
Since:
:$k_1 \eqclass {y_1} {\Img T} + \cdots + k_m \eqclass {y_m} {\Img T} = 0$
we have:
:$k_1 = \cdots = k_m = 0$
Then we also have:
:$\map T x = 0$
As $T$ is injective, it follows that:
:$x = 0$
So, we have shown the injectivity of $\hat T$.
That is, $\hat T$ is a bijective bounded linear transformation.
By Banach Isomorphism Theorem, $\hat T$ is a linear isomorphism.
Now the claim follows from:
:$\Img T = \hat T \sqbrk {X \times \set { {\mathbf 0}_{\Bbb F^m} } }$
Category:Fredholm Operators
Category:Functional Analysis
Image of Generating Set of Vector Space under Linear Transformation is Generating Set of Image

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $T : X \to Y$ be a linear transformation.
Let $G$ be a generating set for $X$. 
Then $T \sqbrk G$ be a generating set for $T \sqbrk X$. 
== Proof ==
Let $y \in T \sqbrk X$.
Then there exists $x \in X$ such that $y = T x$.
Since $G$ generates $X$, there exists $n \in \N$, $x_1, \ldots, x_n \in G$ and $\alpha_1, \ldots, \alpha_n \in K$ such that: 
:$\ds x = \sum_{i \mathop = 1}^n \alpha_i x_i$ 
Then from the linearity of $T$ we have: 
:$\ds y = T x = \sum_{i \mathop = 1}^n \alpha_i T x_i$ 
with $T x_i \in T \sqbrk G$ for each $1 \le i \le n$. 
So:
:$T \sqbrk X \subseteq \map \span {T \sqbrk G}$
and so:
:$T \sqbrk X = \map \span {T \sqbrk G}$
So $T \sqbrk G$ is a generating set for $T \sqbrk X$. 
Category:Linear Transformations
Image of Group Homomorphism is Hausdorff Implies Kernel is Closed

== Theorem ==
Let $G$ and $H$ be topological groups.
Let $f: G \to H$ be a morphism.
Let its image $\Img f$ be Hausdorff.
Then its kernel $\map \ker f$ is closed in $G$.
== Proof ==
By Image of Group Homomorphism is Subgroup, $\Img f$ is a group.
Let $e$ be the identity of $H$.
By Topological Group is Hausdorff iff Identity is Closed, $\set e$ is closed in $\Img f$.
Because $f$ is continuous, $\map \ker f = \map {f^{-1} } e$ is closed in $G$.
Category:Topological Groups
Category:Hausdorff Spaces
Image of Group Homomorphism is Subgroup

== Theorem ==
Let $\phi: G_1 \to G_2$ be a group homomorphism.
Then:
:$\Img \phi \le G_2$
where $\le$ denotes the relation of being a subgroup.
== Proof ==
This is a special case of Group Homomorphism Preserves Subgroups, where we set $H = G_1$.
== Sources ==
* : Chapter $\text{II}$: Groups: Morphisms
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 66$
* : $\S 7$: Homomorphisms and quotient algebras
* : $\S 1.2$
* : $\S 47.4$ Homomorphisms and their elementary properties
* : Chapter $8$: The Homomorphism Theorem: Theorem $8.13: \ (2)$
Category:Subgroups
Category:Group Homomorphisms
Image of Group of Units in Unital Algebra under Unital Algebra Homomorphism

== Theorem ==
Let $R$ be a ring.
Let $A, B$ be unital $R$-algebras.
Let $\phi : A \to B$ be an unital algebra homomorphism.
Let $\map G A$ and $\map G B$ the groups of units of $A$ and $B$ respectively. 
Then:
:$\phi \sqbrk {\map G A} \subseteq \map G B$
=== Corollary ===
== Proof ==
Let ${\mathbf 1}_A$ be the identity element of $A$. 
Let $x \in \map G A$.
Then there exists $y \in A$ such that $x y = y x = {\mathbf 1}_A$. 
We then have $\map \phi {x y} = \map \phi {y x} = \map \phi { {\mathbf 1}_A} = {\mathbf 1}_B$ since $\phi$ is a unital algebra homomorphism.
Since $\phi$ is an algebra homomorphism, we have:
:$\map \phi x \map \phi y = \map \phi y \map \phi x = {\mathbf 1}_B$
Hence $\map \phi y$ is the inverse of $\map \phi x$ in $B$. 
So $\map \phi x \in \map G B$. 
Hence:
:$\phi \sqbrk {\map G A} \subseteq \map G B$
Category:Algebras
Category:Image of Group of Units in Unital Algebra under Unital Algebra Homomorphism