nyuuzyou/wikis huggingface dataset:
Bijection Preserves Set Partition

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection
Let $I$ be an indexing set.
Let $\family {S_i}_{i \mathop \in I}$ be a partitioning of $S$ indexed by $I$.
Hence, let $\set {S_i: i \in I}$ be the resulting partition of $S$.
Then the image of $f$ is a partition of $T$ indexed by $I$ such that:
:$T = \set {f \sqbrk {S_i}: i \in I}$
== Proof ==
By definition of partitioning:
:$(1): \quad \forall i \in I: S_i \ne \O$, that is, none of $S_i$ is empty
:$(2): \quad \ds S = \bigcup_{i \mathop \in I} S_i$, that is, $S$ is the union of $\family {S_i}_{i \mathop \in I}$
:$(3): \quad \forall i, j \in I: i \ne j \implies S_i \cap S_j = \O$, that is, the elements of $\family {S_i}_{i \mathop \in I}$ are pairwise disjoint.
It is sufficient to check that:
:$(1): \quad \forall i \in I: f \sqbrk {S_i} \ne \O$, that is, none of $f \sqbrk {S_i}$ is empty
:$(2): \quad \ds T = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$, that is, $T$ is the union of $\family {f \sqbrk {S_i} }_{i \mathop \in I}$
:$(3): \quad \forall i, j \in I: i \ne j \implies f \sqbrk {S_i} \cap f \sqbrk {S_j} = \O$, that is, the elements of $\family {f \sqbrk {S_i} }_{i \mathop \in I}$ are pairwise disjoint.
We have that:
:$f \sqbrk {S_i} = \set {t \in T: t = \map f s: s \in S_i}$
As $S_i \ne \O$ from $(1)$ it follows that $f \sqbrk {S_i} \ne \O$.
Hence criterion $(1)$ is fulfilled by $f$.
Because $f$ is a bijection, it is a fortiori a surjection.
Hence:
:$\forall t \in T: \exists s \in S: \map f s = t$
But as $S = \set {S_i: i \in I}$ is partitioned by $I$, it follows that:
:$\exists i \in I: s \in S_i$
Hence:
:$\exists i \in I: t \in f \sqbrk {S_i}$
So every element of $T$ is an element of (at least) one of the $f \sqbrk {S_i}$.
That is:
:$\ds T = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$
Hence criterion $(2)$ is fulfilled by $f$.
Because $f$ is a bijection, it is a fortiori an injection.
From Image of Intersection under Injection:
:$\forall S_i, S_j \subseteq S: f \sqbrk {S_i \cap S_j} = f \sqbrk {S_i} \cap f \sqbrk {S_j}$
But for $i \ne j$ we have that $S_i \cap S_j = \O$.
Hence:
:$\forall i, j \in I: i \ne j \implies f \sqbrk {S_i} \cap f \sqbrk {S_j} = \O$
and it follows that criterion $(3)$ is fulfilled by $f$.
Thus all $3$ criteria for $T = \set {f \sqbrk {S_i}: i \in I}$ being a partition are fulfilled.
Category:Set Partitions
Category:Bijections
Bijection between Integers and Even Integers

== Theorem ==
Let $\Z$ be the set of integers.
Let $2 \Z$ be the set of even integers.
Then there exists a bijection $f: \Z \to 2 \Z$ between the two.
== Proof ==
Let $f: \Z \to 2 \Z$ be the mapping defined as:
:$\forall n \in \Z: \map f n = 2 n$
Let $m, n \in \Z$ such that $\map f m = \map f n$.
Thus $f$ is an injection.
Let $n \in 2 Z$.
Then by definition:
:$n = 2 r$
for some $r \in \Z$
That is:
:$n = \map f r$
and so $f$ is a surjection.
So $f$ is both an injection and a surjection.
Hence by definition, $f$ is a bijection.
Category:Examples of Bijections
Category:Integers
Category:Even Integers
Bijection between Power Set of Disjoint Union and Cartesian Product of Power Sets

== Theorem ==
Let $S$ and $T$ be disjoint sets.
Let $\powerset S$ denote the power set of $S$.
Then there exists a bijection between $\powerset {S \cup T}$ and $\paren {\powerset S} \times \paren {\powerset T}$.
Hence:
:$\powerset {S \cup T} \sim \paren {\powerset S} \times \paren {\powerset T}$
where $\sim$ denotes set equivalence.
== Proof ==
Let $\phi: \paren {\powerset S} \times \paren {\powerset T} \to \powerset {S \cup T}$ be defined as:
:$\forall \tuple {A, B} \in \paren {\powerset S} \times \paren {\powerset T}: \map \phi {A, B} = A \cup B$
In order to show that $\phi$ is a bijection, it needs to be shown that $\phi$ has the following properties:
:$\phi$ is a mapping, that is:
::$\phi$ is left-total
::$\phi$ is many-to-one
:$\phi$ is a surjection
:$\phi$ is an injection.
Let $A \subseteq S$ and $B \subseteq T$.
Then:
Thus:
:$\forall A \subseteq S, B \subseteq T: \tuple {A, B} \in \Dom \phi$
and so $\phi$ is left-total.
Let $A_1, A_2 \subseteq S$ and $B_1, B_2 \subseteq T$.
Then:
, suppose $A_1 \cup B_1 \nsubseteq A_2 \cup B_2$.
Then:
That is:
:$\map \phi {A_1, B_1} \ne \map \phi {A_2, B_2} \implies \tuple {A_1, B_1} \ne \tuple {A_2, B_2}$
demonstrating that $\phi$ is many-to-one.
Thus $\phi$ has been shown to be a mapping.
Let $A \cup B \in \powerset {S \cup T}$.
Then:
:$\tuple {A, B} \in \paren {\powerset S} \times \paren {\powerset T}$
by definition of power set and Cartesian product.
That is, $\phi$ is a surjection by definition.
It remains to be shown that $\phi$ is an injection.
Let $\tuple {A_1, B_1} \ne \tuple {A_2, B_2}$.
Then either $A_1 \ne A_2$ or $B_1 \ne B_2$.
That is, at least one of the following holds:
, suppose $A_1 \nsubseteq A_2$.
Then:
:$\exists x \in A_1: x \notin A_2$
and so by Set is Subset of Union:
:$x \in A_1 \cup B_1$
But we have that $S$ and $T$ are disjoint.
That is:
:$x \in A_1 \implies x \notin B_2$
Thus:
:$\exists x \in A_1 \cup B_1: x \notin A_2 \cup B_2$
and so:
:$A_1 \cup B_1 \ne A_2 \cup B_2$
That is:
:$\map \phi {A_1, B_1} \ne \map \phi {A_2, B_2}$
demonstrating that $\phi$ is an injection.
Hence the result.
== Sources ==
* : Chapter $3$. Mappings: Exercise $11$
* : Chapter $1$: Mappings: $\S 14 \alpha$
Category:Power Set
Category:Set Union
Category:Cartesian Product
Bijection between Power Set of nth Initial Section and Initial Section of nth Power of 2

== Theorem ==
Let $\N_n$ be used to denote the first $n$ non-zero natural numbers:
:$\N_n = \set {1, 2, \ldots, n}$
Then there exists a bijection between the power set of $\N_n$ and $\N_{2^n}$.
== Proof ==
Let $\phi: \powerset {\N_n} \to \N_{2^n}$ be defined as:
:$\forall A \in \powerset {\N_n}: \map \phi A = \begin{cases} \ds \sum_{k \mathop \in A} 2^{k - 1} & : A \ne \O \\ 2^k & : A = \O \end{cases}$
Apart from $\O$, every $A \in \powerset {\N_n}$ consists of a set of integers between $1$ and $n$.
The expression $\ds \sum_{k \mathop \in A} 2^{k - 1}$ is the summation of a set of powers of $2$ between $2^0$ and $2^{n - 1}$.
Hence $A$ is seen to be left-total and many-to-one and so is a mapping.
By the Basis Representation Theorem, every integer between $1$ and $2^n - 1$ can be expressed uniquely as the sum of powers of $2$ between $2^0$ and $2^{n - 1}$.
The final integer $2^n$ is mapped to from $\O$.
Thus it is seen that $\phi$ is both an injection and a surjection.
Hence the result.
== Sources ==
* : Chapter $1$: Mappings: $\S 14 \beta$
Category:Power Set
Category:Integer Powers
Bijection between Prüfer Sequences and Labeled Trees

== Theorem ==
There is a one-to-one correspondence between Prüfer sequences and labeled trees.
That is, every labeled tree has a unique Prüfer sequence that defines it, and every Prüfer sequence defines just one labeled tree.
== Proof ==
Let $T$ be the set of all labeled trees of order $n$.
Let $P$ be the set of all Prüfer sequence of length $n-2$.
Let $\phi: T \to P$ be the mapping that maps each tree to its Prüfer sequence.
* From Prüfer Sequence from Labeled Tree, $\phi$ is clearly well-defined, as every element of $T$ maps uniquely to an element of $P$.
* However, from Labeled Tree from Prüfer Sequence, $\phi^{-1}: P \to T$ is also clearly well-defined, as every element of $P$ maps to a unique element of $T$.
Hence the result.
Category:Tree Theory
Category:Combinatorics
Bijection between R x (S x T) and (R x S) x T

== Theorem ==
Let $R$, $S$ and $T$ be sets.
Let $S \times T$ be the Cartesian product of $S$ and $T$.
Then there exists a bijection from $R \times \paren {S \times T}$ to $\paren {R \times S} \times T$.
Hence:
:$\card {R \times \paren {S \times T} } = \card {\paren {R \times S} \times T}$
== Proof ==
Let $\phi: R \times \paren {S \times T} \to \paren {R \times S} \times T$ be the mapping defined as:
:$\forall \tuple {r, \tuple {s, t} } \in R \times \paren {S \times T}: \map \phi {s, t} = \tuple {\tuple {r, s}, t}$
Then $\phi$ is the bijection required, as follows:
The domain of $\phi$ is $R \times \paren {S \times T}$.
Let $\tuple {\tuple {r, s}, t} \in \paren {R \times S} \times T$.
Then there exists $\tuple {r, \tuple {s, t} } \in R \times \paren {S \times T}$ such that $\map \phi {r, \tuple {s, t} } = \tuple {\tuple {r, s}, t}$.
Thus $\phi$ is a surjection.
Let $\map \phi {r_1, \tuple {s_1, t_1} } = \map \phi {r_2, \tuple {s_2, t_2} }$ for some $\tuple {r_1, \tuple {s_1, t_1} }$ and $\tuple {r_2, \tuple {s_2, t_2} }$ in $R \times \paren {S \times T}$.
Then:
and so $\phi$ is an injection.
Hence the result by definition of bijection.
== Sources ==
* : Chapter $3$. Mappings: Exercise $9 \ \text {(ii)}$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{P}$
* : Chapter $1$: Mappings: $\S 13 \gamma$
Category:Cartesian Product
Bijection between S x T and T x S

== Theorem ==
Let $S$ and $T$ be sets.
Let $S \times T$ be the Cartesian product of $S$ and $T$.
Then there exists a bijection from $S \times T$ to $T \times S$.
== Proof ==
Let $\phi: S \times T \to T \times S$ be the mapping defined as:
:$\forall \tuple {s, t} \in S \times T: \map \phi {s, t} = \tuple {t, s}$
Then $\phi$ is the bijection required, as follows:
The domain of $\phi$ is $S \times T$.
Let $\tuple {t, s} \in T \times S$.
Then there exists $\tuple {s, t} \in S \times T$ such that $\map \phi {s, t} = \tuple {t, s}$.
Thus $\phi$ is a surjection.
Let $\map \phi {s_1, t_1} = \map \phi {s_2, t_2}$ for some $\tuple {s_1, t_1}$ and $\tuple {s_2, t_2}$ in $S \times T$.
Then:
and so $\phi$ is an injection.
Hence the result by definition of bijection.
== Sources ==
* : Chapter $1$: Mappings: $\S 13 \beta$
Category:Cartesian Product
Bijection between Specific Elements

== Lemma ==
Let $A$ and $B$ be sets.
Let $f: A \leftrightarrow B$ be a bijection.
Then given $a \in A$ and $b \in B$, there exists another bijection $g: A \leftrightarrow B$ such that $\map g a = b$.
== Proof ==
Let $\map f a = b'$.
Since $f$ is surjective, there is $a' \in A$ such that $\map f {a'} = b$.
We define the mapping $g: A \to B$ as:
:$\map g x = \begin {cases}
b & : x = a \\
b' & : x = a' \\
\map f x & : x \in A \setminus \set {a, a'} \end {cases}$
=== $g$ is Surjective ===
Let $y \in B \setminus \set {b, b'}$.
Then:
:$\exists x \in A: \map f x = y$
and so by definition of $g$:
:$\exists x \in A: \map g x = y$
Otherwise:
:$y = b \implies b = \map g a$
:$y = b' \implies b = \map g {a'}$
and so $y$ is seen to be surjective.
=== $g$ is Injective ===
Suppose $\map g x = \map g {x'}$.
If $\map g x, \map g {x'} \in B \setminus \set {b, b'}$ then $\map f x = \map f {x'}$.
As $f$ is bijective and so injective it follows that $x = x'$.
If $\map g x, \map g {x'} \in \set {b, b'}$ then either:
:$\map g x = \map g {x'} = b$ and so $x = x' = a$
or
:$\map g x = \map g {x'} = b'$ and so $x = x' = a'$
For the other two cases:
:$\map g x \in \set {b, b'}, \map g {x'} \notin \set {b, b'}$
:$\map g x \notin \set {b, b'}, \map g {x'} \in \set {b, b'}$
it is clear that $\map g x \ne \map g {x'}$.
So $g$ has been shown to be injective.
=== $g$ is Bijective ===
So $g$ is both injective and surjective, and so by definition a bijection.
== Motivation ==
What this means is that given any two equivalent sets, it is always possible to find a bijection that connects any given pair of elements, one in one set and one in the other.
Category:Bijections
Bijection from Cartesian Product of Initial Segments to Initial Segment

== Theorem ==
Let $\N_k$ be used to denote the set of the first $k$ non-zero natural numbers:
:$\N_k := \set {1, 2, \ldots, k}$
Then a bijection can be established between $\N_k \times \N_l$ and $\N_{k l}$, where $\N_k \times \N_l$ denotes the Cartesian product of $\N_k$ and $\N_l$.
== Proof ==
Let $\phi: \N_k \times \N_l \to \N_{k l}$ be defined as:
:$\forall \tuple {m, n} \in \N_k \times \N_l: \map \phi {m, n} = \paren {m - 1} \times l + n$
First it is confirmed that the codomain of $\phi$ is indeed $\N_{k l}$.
== Sources ==
* : Chapter $1$: Mappings: $\S 13 \alpha$
Category:Bijections
Category:Natural Numbers
Bijection from Divisors to Subgroups of Cyclic Group

== Theorem ==
Let $G$ be a cyclic group of order $n$ generated by $a$.
Let $S = \set {m \in \Z_{>0}: m \divides n}$ be the set of all divisors of $n$.
Let $T$ be the set of all subgroups of $G$
Let $\phi: S \to T$ be the mapping defined as:
:$\phi: m \to \gen {a^{n / m} }$
where $\gen {a^{n / m} }$ is the subgroup generated by $a^{n / m}$.
Then $\phi$ is a bijection.
== Proof ==
From Subgroup of Finite Cyclic Group is Determined by Order, there exists exactly one subgroup $\gen {a^{n / m} }$ of $G$ with $a$ elements.
So the mapping as defined is indeed a bijection by definition.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.8$
Category:Subgroups
Category:Cyclic Groups
Bijection has Left and Right Inverse

== Theorem ==
Let $f: S \to T$ be a bijection.
Let:
: $I_S$ be the identity mapping on $S$
: $I_T$ be the identity mapping on $T$.
Let $f^{-1}$ be the inverse of $f$.
Then:
: $f^{-1} \circ f = I_S$
and:
: $f \circ f^{-1} = I_T$
where $\circ$ denotes composition of mappings.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Inverse Mappings
Category:Bijections
Category:Bijection has Left and Right Inverse
Bijection has Left and Right Inverse/Proof 1

== Theorem ==
== Proof ==
Let $f$ be a bijection.
Then it is both an injection and a surjection, thus both the described $g_1$ and $g_2$ must exist from Injection iff Left Inverse and Surjection iff Right Inverse.
The fact that $g_1 = g_2 = f^{-1}$ follows from Left and Right Inverses of Mapping are Inverse Mapping.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.9$
Category:Bijection has Left and Right Inverse
Bijection has Left and Right Inverse/Proof 2

== Theorem ==
== Proof ==
Suppose $f$ is a bijection.
From Bijection iff Inverse is Bijection and Composite of Bijection with Inverse is Identity Mapping, it is shown that the inverse mapping $f^{-1}$ such that:
: $f^{-1} \circ f = I_S$
: $f \circ f^{-1} = I_T$
is a bijection.
Category:Bijection has Left and Right Inverse
Bijection has Left and Right Inverse/Proof 3

== Theorem ==
== Proof ==
Let $f$ be a bijection.
By definition, $f$ is a mapping, and hence also by definition a relation.
Hence the result Bijective Relation has Left and Right Inverse applies directly and so:
:$f^{-1} \circ f = I_S$
and
:$f \circ f^{-1} = I_T$
Category:Bijection has Left and Right Inverse
Bijection iff Left and Right Cancellable

== Theorem ==
Let $f$ be a mapping.
Then $f$ is a bijection $f$ is both left cancellable and right cancellable.
== Proof ==
Follows directly from:
:Injection iff Left Cancellable: $f$ is an injection $f$ is left cancellable
:Surjection iff Right Cancellable: $f$ is a surjection $f$ is right cancellable.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.9$
Category:Bijections
Bijection iff Left and Right Inverse

== Theorem ==
Let $f: S \to T$ be a mapping.
$f$ is a bijection :
where:
:$g_1$ and $g_2$ are mappings
:$\circ$ denotes composition of mappings.
It also follows that it is necessarily the case that $g_1 = g_2$ for such to be possible.
=== Corollary ===
== Proof ==
=== Necessary Condition ===
Let $f: S \to T$ be a mapping.
Let $f$ be such that:
where both $g_1$ and $g_2$ are mappings.
Then from Left and Right Inverse Mappings Implies Bijection it follows that $f$ is a bijection.
From Left and Right Inverses of Mapping are Inverse Mapping it follows that:
:$g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the inverse of $f$.
=== Sufficient Condition ===
Let $f: S \to T$ be a bijection.
Then from Bijection has Left and Right Inverse it follows that:
:$f^{-1} \circ f = I_S$
and:
:$f \circ f^{-1} = I_T$
where $f^{-1}$ is the inverse of $f$.
== Also see ==
* Composite of Bijection with Inverse is Identity Mapping for the converse of this result.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions: Theorem $9$
* : $\S 2$: Sets and functions: Inverse images and inverse functions
* 
* : $1$: Set Theory and Logic: $\S 2$: Functions: Lemma $2.1$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.5 \ \text{(e)}$
* : Appendix $\text{A}$: Set Theory: Bijections
* 
Category:Bijections
Category:Inverse Mappings
Bijection iff Left and Right Inverse/Corollary

== Corollary to Bijection iff Left and Right Inverse ==
Let $f: S \to T$ and $g: T \to S$ be mappings such that:
Then both $f$ and $g$ are bijections.
== Proof ==
Suppose we have such mappings $f$ and $g$ with the given properties.
From Bijection iff Left and Right Inverse, we have that $f$ is a bijection, by considering $g = g_1$ and $g = g_2$.
It directly follows that by setting $g = f, f = g_1, f = g_2$, the result Bijection iff Left and Right Inverse can be used the other way about.
== Sources ==
* : Introduction $\S 2$: Product sets, mappings
* : $\S 25.1$: Some further results and examples on mappings
Category:Bijections
Category:Inverse Mappings
Bijection iff exists Mapping which is Left and Right Inverse

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then $f$ is a bijection :
:there exists a mapping $g: T \to S$ such that:
::$g \circ f = I_S$
::$f \circ g = I_T$
:where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.
== Proof ==
=== Necessary Condition ===
Let $f$ be a bijection.
Then for each $y \in T$ there exists one and only one $x \in S$ such that $\map f x = y$.
That is, that there exists a mapping $g: T \to S$ with the property that:
: $\forall y \in T: \exists x \in S: \map g y = x$
Let $y \in T$.
Let $x = g \map g y$.
Then:
and:
demonstrating that $g$ has the property that:
:$g \circ f = I_S$
:$f \circ g = I_T$
=== Sufficient Condition ===
Suppose there exists a $g$ which satisfies the conditions on $f$.
By Condition for Composite Mapping to be Identity:
:$(1): \quad$ from $g \circ f = I_S$ it follows that $f$ is an injection.
:$(2): \quad$ from $f \circ g = I_T$ it follows that $f$ is a surjection.
Thus, by definition, $f$ is a bijection.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Theorem $2$
Category:Bijections
Bijection in Yoneda Lemma for Covariant Functors

== Theorem ==
Let $C$ be a locally small category.
Let $\mathbf {Set}$ be the category of sets.
Let $F: C \to \mathbf {Set}$ be a covariant functor.
Let $A \in C$ be an object.
Let $I_A$ be its identity morphism.
Let $h^A = \map {\operatorname {Hom} } {A, -}$ be its covariant hom-functor.
The class of natural transformations $\map {\operatorname {Nat} } {h^A, F}$ is a small class, and:
:$\alpha: \map {\operatorname {Nat} } {h^A, F} \to \map F A: \eta \mapsto \map {\eta_A} {I_A}$
:$\beta: \map F A \to \map {\operatorname {Nat} } {h^A, F}: u \mapsto \paren {X \mapsto \paren {f \mapsto \map {\paren {\map F f} } u} }$
are inverses of each other.
== Outline of proof ==
The crucial point is that the role of morphisms $f \in \map {h^A} B = \map {\operatorname {Hom} } {A, B}$ is close to their role under the hom functor $h^A$: We have $f = f \circ I_A = \map {\paren {\map {h^A} f} } {I_A}$.
Natural transformations $\eta: h^A \to F$ translate what happens with $f$ under $h^A$ to something involving only $F$, so that $\eta$ is determined by $\map {\eta_A} {I_A} = \map \alpha \eta$.
Making the calculation explicit, we know what the reverse bijection $\beta$ should be.
== Proof ==
The fact that $\map {\operatorname {Nat} } {h^A, F}$ is a small class follows when we prove that the mappings are bijections.
=== Injectivity ===
Let $\eta \in \map {\operatorname {Nat} } {h^A, F}$ and $B \in C$ an object.
Let $f \in \map {h^A} B = \map {\operatorname {Hom} } {A, B}$ be a morphism.
Then:
which shows that $\eta$ is determined by $\map {\eta_A} {I_A} = \map \alpha \eta$.
=== Surjectivity ===
Let $u \in \map F A$.
Inspired by the result above, we define $\map \beta u = \eta$ by $\map {\eta_B} f = \map {\map F f} u$ for $B \in C$ and $f \in \map {\operatorname {Hom} } {A, B}$.
Let $B, D \in C$ and $g: B \to D$ be a morphism.
We have to show that the following diagram commutes:
:$\xymatrix{
\map {\operatorname {Hom} } {A, B} \ar[d]^{\eta_B} \ar[r]^{g_*} & \map {\operatorname {Hom} } {A, D} \ar[d]^{\eta_D} \\
F(B) \ar[r]^{\map F g} & \map F D}$
where $g_* = \map {\operatorname {Hom} } {A, g}$ is the postcomposition mapping.
Let $f \in \map {\operatorname {Hom} } {A, B}$.
We have:
and:
=== Reverse bijections ===
By construction, $\map \beta {\map \alpha \eta} = \eta$ for all $\eta \in \map {\operatorname {Nat} } {h^A, F}$.
It remains to show that $\map \alpha {\map \beta u} = u$ for all $u \in \map F A$.
Let $\eta = \map \beta u$.
Then
== Also see ==
* Bijection in Yoneda Lemma for Contravariant Functors
Category:Category Theory
Bijection is Open iff Closed

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a bijection.
Then $f$ is open $f$ is closed.
== Proof ==
Let $f$ be a bijection.
Suppose $f$ is an open mapping.
From the definition of open mapping:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
As $f$ is a bijection:
:$f \sqbrk {S_1 \setminus H} = f \sqbrk {S_1} \setminus f \sqbrk H = S_2 \setminus f \sqbrk H$
By definition of closed set:
:$S_1 \setminus H$ is closed in $T_1$
and as $f$ is an open mapping:
:$f \sqbrk {S_1 \setminus H} = S_2 \setminus f \sqbrk H$
is closed in $T_2$.
Hence by definition $f$ is a closed mapping.
A similar argument demonstrates that if $f$ is closed then it is open.
== Also see ==
* Bijection is Open iff Inverse is Continuous
* Open Mapping is not necessarily Closed Mapping
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Open Mappings
Category:Closed Mappings
Category:Bijections
Bijection is Open iff Inverse is Continuous

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a bijection.
Then $f$ is open $f^{-1}$ is continuous.
== Proof ==
Let $f$ be a bijection.
Let $g := f^{-1}$.
By Bijection iff Inverse is Bijection we have that $g$ is a bijection and that $g^{-1} = f$.
Let $f$ be open.
Then by definition of open mapping:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
taking $H \in \tau_1$ by definition of open in $T_1$.
But $f = g^{-1}$ and so:
:$\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$
which is exactly the definition for $g$ to be continuous.
The argument works the other way.
Let $g$ be continuous.
Then by definition of continuous mapping:
:$\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$
But $g^{-1} = f$ and so:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
which is exactly the definition for $f$ to be open.
== Also see ==
* Bijection is Open iff Closed
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Open Mappings
Category:Continuous Mappings
Bijection on Total Ordering is Order Isomorphism

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $f: S \to T$ be a bijection to an arbitrary set $T$.
Let $\RR$ be a relation on $T$ defined such that:
:$\RR: = \set {\tuple {\map f x, \map f y}: x \preccurlyeq y}$
Then $f$ is an order isomorphism between $\struct {S, \preccurlyeq}$ and $\struct {T, \RR}$.
== Proof ==
From Bijection on Total Ordering reflects Total Ordering, we have that $\RR$ is a total ordering.
We have that $f$ is a bijection.
Hence a fortiori $f$ is also a surjection.
Let $x, y \in S$ such that $x \preccurlyeq y$.
Then by definition of $\RR$:
:$\map f x \mathrel \RR \map f y$
Now let $\map f x \mathrel \RR \map f y$.
Then again by definition of $\RR$:
:$x \preccurlyeq y$
That is, $f$ is by definition an order isomorphism between $\struct {S, \preccurlyeq}$ and $\struct {T, \RR}$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $41 \ \text {(d)}$
Category:Total Orderings
Category:Bijections
Category:Order Isomorphisms