hcju/proofwikiqa huggingface dataset:
Let $\mathcal R_1$ and $\mathcal R_2$ be equivalence relations on $S$. Let $\mathcal R_3 = \mathcal R_1 \cap \mathcal R_2$. Checking in turn each of the criteria for equivalence: === Reflexive === Equivalence relations are by definition reflexive. So, by Intersection of Reflexive Relations is Reflexive, so is $\mathcal R_3$. === Symmetric === Equivalence relations are by definition symmetric. So, by Intersection of Symmetric Relations is Symmetric, so is $\mathcal R_3$. === Transitive === Equivalence relations are by definition transitive. So, by Intersection of Transitive Relations is Transitive, so is $\mathcal R_3$. Thus $\mathcal R_3$ is an equivalence. {{qed}}
This can be shown by giving an example. Let $S = \left\{{a, b, c}\right\}$, and let $\mathcal R_1$ and $\mathcal R_2$ be equivalences on $S$ such that: : $\left[\!\left[{a}\right]\!\right]_{\mathcal R_1} = \left[\!\left[{b}\right]\!\right]_{\mathcal R_1} = \left\{{a, b}\right\}$ : $\left[\!\left[{c}\right]\!\right]_{\mathcal R_1} = \left\{{c}\right\}$ : $\left[\!\left[{a}\right]\!\right]_{\mathcal R_2} = \left\{{a}\right\}$ : $\left[\!\left[{b}\right]\!\right]_{\mathcal R_2} = \left[\!\left[{c}\right]\!\right]_{\mathcal R_2} = \left\{{b, c}\right\}$ Let $\mathcal R_3 = \mathcal R_1 \cup \mathcal R_2$. Then: {{begin-eqn}} {{eqn | o =        | r = \left({a, b}\right) \in \mathcal R_1       | c =  }} {{eqn | o =        | r = \left({b, c}\right) \in \mathcal R_2       | c =  }} {{eqn | o = \implies       | r = \left({a, b}\right) \in \mathcal R_1 \cup \mathcal R_2 \land \left({b, c}\right) \in \mathcal R_1 \cup \mathcal R_2       | c =  }} {{eqn | o = \implies       | r = \left({a, b}\right) \in \mathcal R_3 \land \left({b, c}\right) \in \mathcal R_3       | c =  }} {{end-eqn}} However: {{begin-eqn}} {{eqn | o =        | r = \left({a, c}\right) \notin \mathcal R_1       | c =  }} {{eqn | o =        | r = \left({a, c}\right) \notin \mathcal R_2       | c =  }} {{eqn | o = \implies       | r = \left({a, c}\right) \notin \mathcal R_1 \cup \mathcal R_2       | c =  }} {{eqn | o = \implies       | r = \left({a, c}\right) \notin \mathcal R_3       | c =  }} {{end-eqn}} So $\mathcal R_3$ is not transitive, and therefore $\mathcal R_3 = \mathcal R_1 \cup \mathcal R_2$ is not an equivalence relation. {{qed}}
We have that the Union of Reflexive Relations is Reflexive. We also have that the Union of Symmetric Relations is Symmetric. However, we also have that the Union of Transitive Relations Not Always Transitive. Hence the union of two equivalence relations is not always an equivalence relation. {{qed}}
=== Definition 1 implies Definition 2 === Let $\RR$ be an equivalence relation by definition 1. By definition, $\RR$ is reflexive, symmetric and transitive. From Relation Contains Diagonal Relation iff Reflexive: :$\Delta_S \subseteq \RR$ From Relation equals Inverse iff Symmetric: :$\RR^{-1} = \RR$ and so by definition of set equality: :$\RR^{-1} \subseteq \RR$ From Relation contains Composite with Self iff Transitive: :$\RR \circ \RR \subseteq \RR$ From Union is Smallest Superset: :$\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$ Thus $\RR$ is an equivalence relation by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\RR$ be an equivalence relation by definition 2. That is: :$\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$ By Union is Smallest Superset: :$(1): \quad \Delta_S \subseteq \RR$ :$(2): \quad \RR^{-1} \subseteq \RR$ :$(3): \quad \RR \circ \RR \subseteq \RR$ From Relation Contains Diagonal Relation iff Reflexive, $(1)$ gives directly that $\RR$ is reflexive. From Inverse Relation Equal iff Subset, $(2)$ gives that $\RR^{-1} = \RR$. So from Relation equals Inverse iff Symmetric $\RR$ is symmetric. From Relation contains Composite with Self iff Transitive, $(3)$ gives that $\RR$ is transitive. So $\RR$ has been shown to be reflexive, symmetric and transitive. Thus $\RR$ is an equivalence relation by definition 1. {{qed}}
=== Necessary Condition === Let $\RR$ be an equivalence relation. Then by definition, it is reflexive, symmetric and transitive. As $\RR$ is reflexive, we have $\Delta_S \subseteq \RR$ from Relation Contains Diagonal Relation iff Reflexive. As $\RR$ is transitive: :$\RR \circ \RR \subseteq \RR$ from Relation contains Composite with Self iff Transitive. But as $\RR$ is symmetric: :$\RR = \RR^{-1}$ from Relation equals Inverse iff Symmetric. Thus: :$\RR \circ \RR^{-1} \subseteq \RR$ Now we need to show that $\RR \subseteq \RR \circ \RR^{-1}$: Let $\tuple {x, y} \in \RR$. Then as $\RR$ is reflexive: :$\tuple {x, y} \in \RR \land \tuple {y, y} \in \RR$ and so: :$\tuple {x, y} \in \RR \circ \RR$ As $\RR = \RR^{-1}$, it follows that: :$\RR \subseteq \RR \circ \RR^{-1}$ So it has been shown: :$\Delta_S \subseteq \RR$ and $\RR = \RR \circ \RR^{-1}$ {{qed|lemma}} === Sufficient Condition === Now, let $\Delta_S \subseteq \RR$ and $\RR = \RR \circ \RR^{-1}$. From Relation Contains Diagonal Relation iff Reflexive, $\RR$ is reflexive. Let $\tuple {x, y} \in \RR$. Then: :$\tuple {x, y} \in \RR \circ \RR^{-1}$ So: :$\exists z \in S: \tuple {x, z} \in \RR, \tuple {z, y} \in \RR^{-1}$ and so: :$\tuple {y, z} \in \RR$ That is, by definition, $\RR$ is transitive. As $\tuple {x, z} \in \RR$: :$\tuple {z, x} \in \RR^{-1}$ and so: :$\tuple {y, x} \in \RR$ So it follows that $\RR$ is symmetric. Thus $\RR$ is reflexive, symmetric and transitive. Therefore, by definition, $\RR$ is an equivalence relation. {{qed}}
=== Necessary Condition === Let: :$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f$ Then: {{begin-eqn}} {{eqn | o =       | r = \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f       | c = }} {{eqn | o = \leadsto       | r = \map f {x_1} = y \land \map f {x_2} = y       | c = {{Defof|Mapping}} }} {{eqn | o = \leadsto       | r = \map f {x_1} = \map f {x_2}       | c = Equality is Equivalence Relation }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let: :$\map f {x_1} = \map f {x_2}$ Then: {{begin-eqn}} {{eqn | o =       | r = \exists y \in \Rng f: \map f {x_1} = y = \map f {x_2}       | c = }} {{eqn | o = \leadsto       | r = \exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f       | c = {{Defof|Mapping}} }} {{end-eqn}} {{qed|lemma}} The result follows from the definition of logical equivalence. {{qed}} Category:Mapping Theory r2h6pslsz6g6qeely8104vu5cthaxa1
We have that $f$ is a mapping. Hence $f$ is a fortiori a many-to-one relation. Then from Inverse of Many-to-One Relation is One-to-Many, $f^{-1}$ is one-to-many. {{Qed}} Category:Inverse Relations Category:Inverse Mappings ayh6r7f3eifj9ml49b63dzgczo2bazq
=== Necessary Condition === Let $f_c: S \to T$ be a constant mapping. Then from the definition of the image of a element: :$\forall x \in S: \map {f_c} x = c \implies \Img S = \set c$ Thus the image of $f_c: S \to T$ is a singleton. === Sufficient Condition === Let $f: S \to T$ such that its image is a singleton $\set c \subseteq T$. {{begin-eqn}} {{eqn | l = \forall x \in S: \map f x       | o = \in       | r = \Img f       | c = {{Defof|Image of Mapping}} }} {{eqn | ll= \leadsto       | l = \forall x \in S: \map f x       | o = \in       | r = \set c       | c = By Hypothesis }} {{eqn | ll= \leadsto       | l = \forall x \in S: \map f x       | r = c       | c = {{Defof|Singleton}} }} {{end-eqn}} Thus $f: S \to T$ is a constant mapping (and we can write it $f_c: S \to T$). {{qed}}
We use the definition of relation equality, as follows: === Equality of Codomains === The codomains of $\RR$ and $\RR \circ \Delta_S$ are both equal to $T$ from Codomain of Composite Relation. === Equality of Domains === The domains of $\RR$ and $\RR \circ \Delta_S$ are also easily shown to be equal. From Domain of Composite Relation: :$\Dom {\RR \circ \Delta_S} = \Dom {\Delta_S}$ But from the definition of the diagonal relation: :$\Dom {\Delta_S} = \Img {\Delta_S} = S$ === Equality of Relations === The composite of $\Delta_S$ and $\RR$ is defined as: :$\RR \circ \Delta_S = \set {\tuple {x, z} \in S \times T: \exists y \in S: \tuple {x, y} \in \Delta_S \land \tuple {y, z} \in \RR}$ But by definition of the diagonal relation on $S$, we have that: :$\tuple {x, y} \in \Delta_S \implies x = y$ Hence: :$\RR \circ \Delta_S = \set {\tuple {y, z} \in S \times T: \exists y \in S: \tuple {y, y} \in \Delta_S \land \tuple {y, z} \in \RR}$ But as $\forall y \in S: \tuple {y, y} \in \Delta_S$, this means: :$\RR \circ \Delta_S = \set {\tuple {y, z} \in S \times T: \tuple {y, z} \in \RR}$ That is: :$\RR \circ \Delta_S = \RR$ Hence the result. {{qed}}
=== Equality of Domains === The domains of $f$ and $I_T \circ f$ are equal from Domain of Composite Relation: :$\Dom {I_T \circ f} = \Dom f$ {{qed|lemma}} === Equality of Codomains === The codomains of $f$ and $f \circ I_S$ are also easily shown to be equal. From Codomain of Composite Relation, the codomains of $I_T \circ f$ and $I_T$ are both equal to $T$. But from the definition of the identity mapping, the codomain of $I_T$ is $\Dom {I_T} = T$ {{qed|lemma}} === Equality of Relations === The composite of $f$ and $I_T$ is defined as: :$I_T \circ f = \set {\tuple {x, z} \in S \times T: \exists y \in T: \tuple {x, y} \in f \land \tuple {y, z} \in I_T}$ But by definition of the identity mapping on $T$, we have that: :$\tuple {y, z} \in I_T \implies y = z$ Hence: :$I_T \circ f = \set {\tuple {x, y} \in S \times T: \exists y \in T: \tuple {x, y} \in f \land \tuple {y, y} \in I_T}$ But as $\forall y \in T: \tuple {y, y} \in I_T$, this means: :$I_T \circ f = \set {\tuple {x, y} \in S \times T: \tuple {x, y} \in f}$ That is: :$I_T \circ f = f$ {{qed|lemma}} Hence the result, by Equality of Mappings. {{qed}}
By definition, a mapping is also a relation. Also by definition, the identity mapping is the same as the diagonal relation. Thus Diagonal Relation is Left Identity can be applied directly. {{qed}}
=== 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 $y = f \paren x$. 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: :$\forall s \in f^{-1} \paren y: s = x$ Thus: :$\map {f^{-1} } y = \set x$ That is: :$x = \map {f^{-1} } y$ {{qed|lemma}} === Necessary Condition === Let $\map {f^{-1} } y = x$. By definition of inverse mapping: :$\map f x = y$ {{qed}}
From the definition: a mapping $f: Y \to Z$ is left cancellable {{iff}}: :$\forall X: \forall g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2 \implies g_1 = g_2$ === Necessary Condition === {{:Injection iff Left Cancellable/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Injection iff Left Cancellable/Sufficient Condition}}{{Qed}}
Let $f: S \to T$ be an injection by definition 1. Thus: :$\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$ First we note that:  :$t \in \Img f \implies \exists x \in \Dom f: \map f x = t$ thus fulfilling the condition for $f^{-1} {\restriction_{\Img f} }$ to be left-total. Now let: :$t \in \Img f: \tuple {t, y}, \tuple {t, z} \in f^{-1}$ Thus: {{begin-eqn}} {{eqn | l = \tuple {t, y}, \tuple {t, z}       | o = \in       | r = f^{-1} {\restriction_{\Img f} }       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {y, t}, \tuple {z, t}       | o = \in       | r = f       | c = {{Defof|Inverse of Mapping}} }} {{eqn | ll= \leadsto       | l = \map f y = t       | r = \map f z       | c = Equality of Elements in Range of Mapping }} {{eqn | ll= \leadsto       | l = y       | r = z       | c = as $f$ is injective }} {{end-eqn}} So by the definition of mapping, $f^{-1} {\restriction_{\Img f} }$ is a mapping. So $f$ is an injection by definition 3. {{qed|lemma}} Let $f: S \to T$ be an injection by definition 3. Then: :$f^{-1} {\restriction_{\Img f} }: \Img f \to \Dom f$ is a mapping where $f^{-1} {\restriction_{\Img f} }$ is the restriction of the inverse of $f$ to the image set of $f$. We need to show that: :$\forall x, z \in \Dom f: \map f x = \map f z \implies x = z$ So, pick any $x, z \in \Dom f$ such that: :$\map f x = \map f z$ Then: {{begin-eqn}} {{eqn | l = \map f x       | r = \map f z       | c =  }} {{eqn | ll= \leadsto       | l = \exists y \in \Dom f: \tuple {x, y}, \tuple {z, y}       | o = \in       | r = f       | c = {{Defof|Mapping}} }} {{eqn | ll= \leadsto       | l = \tuple {y, x}, \tuple {y, z}       | o = \in       | r = f^{-1} {\restriction_{\Img f} }       | c = {{Defof|Inverse of Mapping}} }} {{eqn | ll= \leadsto       | l = x       | r = z       | c = as it is specified that $f^{-1} {\restriction_{\Img f} }$ is a mapping }} {{end-eqn}} So $f$ is an injection by definition 1. {{Qed}}
From the definition of the identity mapping: :$\forall x \in S: \map {I_S} x = x$ So: :$\map {I_S} x = \map {I_S} y \implies x = y$ So by definition $I_S$ is an injection. {{Qed}}
Let $f$ and $g$ be injections. Then: {{begin-eqn}} {{eqn | l = \map {f \circ g} x       | r = \map {f \circ g} y       | c =  }} {{eqn | ll= \leadsto       | l = \map f {\map g x}       | r = \map f {\map g y}       | c = {{Defof|Composition of Mappings}} }} {{eqn | ll= \leadsto       | l = \map g x       | r = \map g y       | c = as $f$ is injective }} {{eqn | ll= \leadsto       | l = x       | r = y       | c = as $g$ is injective }} {{end-eqn}} {{qed}}
Let $g \circ f$ be injective. We need to show that $\map f a = \map f b \implies a = b$. So suppose $\map f a = \map f b$. Then: {{begin-eqn}} {{eqn | l = \map {g \circ f} a       | r = \map g {\map f a}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map g {\map f b}       | c = by hypothesis }} {{eqn | r = \map {g \circ f} b       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} and so $a = b$ as $g \circ f$ is an injection. So we have shown that: :$\map f a = \map f b \implies a = b$ Hence the result from the definition of injection. {{qed}}
Let: :$\exists g: T \to S: g \circ f = I_S$ From Identity Mapping is Injection, $I_S$ is injective, so $g \circ f$ is injective. So from Injection if Composite is Injection, $f$ is an injection. Note that the existence of such a $g$ ''requires'' that $S \ne \O$. {{qed|lemma}} Now, assume $f$ is an injection. We now define a mapping $g: T \to S$ as follows. As $S \ne \O$, we choose $x_0 \in S$. By definition of injection: :$f^{-1} {\restriction_{\Img f} } \to S$ is a mapping so it is possible to define: :$\map g y = \begin{cases} x_0: & y \in T \setminus \Img f \\ \map {f^{-1} } y: & y \in \Img f \end{cases}$ :750px It does not matter what the elements of $T \setminus \Img f$ are. Using the construction given, the equation $g \circ f = I_S$ holds whatever value (or values) we choose for $g \sqbrk {T \setminus \Img f}$. The remaining elements of $T$ can be mapped arbitrarily, and will not affect the image of any $x \in S$ under the mapping $g \circ f$. So, for all $x \in S$: :$\map {g \circ f} x = \map g {\map f x}$ is the unique element of $S$ which $f$ maps to $\map f x$. This unique element is $x$. Thus $g \circ f = I_S$. {{Qed}}
Take the result Condition for Composite Mapping on Left: Let $A, B, C$ be sets. Let $f: A \to B$ and $g: A \to C$ be mappings. Then: :$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$ {{iff}}: :$\exists h: B \to C$ such that $h$ is a mapping and $h \circ f = g$. Let $C = A = S$, let $B = T$ and let $g = I_S$. Then the above translates into: :$\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$ {{iff}}: :$\forall x, y \in S: \map f x = \map f y \implies \map {I_S} x = \map {I_S} y$ But as $\map {I_S} x = x$ and $\map {I_S} y = y$ by definition of identity mapping, it follows that: :$\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$ {{iff}}: :$\forall x, y \in S: \map f x = \map f y \implies x = y$ which is our result. {{qed}}
Let $f: S \to T$ be an injection. Then $f$ is a one-to-many relation. By Inverse of Many-to-One Relation is One-to-Many, $f^{-1}: T \to S$ is many-to-one. By Many-to-One Relation Extends to Mapping, there is a Mapping $g: T \to S$ such that $f^{-1} \subseteq g$. Let $\tuple {x, y} \in g \circ f$. Then: :$\exists z \in T: \tuple {x, z} \in f, \tuple {z, y} \in g$ Since $\tuple {x, z} \in f$: :$\tuple {z, x} \in f^{-1} \subseteq g$ Since $\tuple {z, y} \in g$, $\tuple {z, x} \in g$ and $g$ is a mapping: :$x = y$ so: :$\tuple {x, y} \in I_S$ So we see that: :$g \circ f \subseteq I_S$ Let $x \in S$. Then: :$\tuple {x, \map f x} \in f$ and: :$\tuple {\map f x, x} \in f^{-1} \subseteq g$ So: :$\tuple {x, x} \in g \circ f$ Thus: :$I_S \subseteq g \circ f$ By definition of set equality: :$I_S = g \circ f$ {{qed}}
Suppose $\map {i_S} {s_1} = \map {i_S} {s_2}$. {{begin-eqn}} {{eqn | l = \map {i_S} {s_1}       | r = s_1       | c = {{Defof|Inclusion Mapping}} }} {{eqn | l = \map {i_S} {s_2}       | r = s_2       | c = {{Defof|Inclusion Mapping}} }} {{eqn | l = \map {i_S} {s_1}       | r = \map {i_S} {s_2}       | c = by definition }} {{eqn | ll= \leadsto       | l = s_1       | r = s_2       | c = from above }} {{end-eqn}} Thus $i_S$ is an injection by definition. {{qed}}
The identity mapping is defined as: :$\forall y \in S: \map {I_S} y = y$ Then we have: {{begin-eqn}} {{eqn | l = \forall y \in S: \exists x \in S: x       | r = y       | c = that is, $y$ itself }} {{eqn | ll= \leadsto       | l = \forall y \in S: \exists x \in S: \map {I_S} x       | r = y       | c = Definition of $I_S$ }} {{end-eqn}} Hence the result. {{qed}}
From Image is Subset of Codomain: :$\Img g \subseteq T$ Furthermore, by definition of image, we have: :$\forall s \in S: g \paren s \in \Img g$ Therefore, $g$ may be viewed as a mapping $g: S \to \Img g$. Thus, by definition, $g$ is a surjection. {{qed}}
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be surjections. Then: {{begin-eqn}} {{eqn | l = \forall z \in S_3: \exists y \in S_2: \map g y       | r = z       | c = {{Defof|Surjection}} }} {{eqn | ll= \leadsto       | l = \exists x \in S_1: \map f x       | r = y       | c = {{Defof|Surjection}} }} {{end-eqn}} By definition of a composite mapping: :$\map {g \circ f} x = \map g {\map f x} = \map g y = z$ Hence $g \circ f$ is surjective. {{qed}}
Let $g \circ f$ be surjective. Fix $z \in S_3$.  Now find an $x \in S_1: \map {g \circ f} x = z$. The surjectivity of $g \circ f$ guarantees this can be done. Now find an $y \in S_2: f \paren x = y$. $f$ is a mapping and therefore a left-total relation; which guarantees this too can be done. It follows that: {{begin-eqn}} {{eqn | l = \map g y       | r = \map g {\map f x} }} {{eqn | r = \map {g \circ f} x       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = z       | c = Choice of $x$ }} {{end-eqn}} {{qed}}
Assume $\exists g: T \to S: f \circ g = I_T$. From Identity Mapping is Surjection, $I_T$ is surjective, so $f \circ g$ is surjective. So from Surjection if Composite is Surjection, $f$ is a surjection. Note that the existence of such a $g$ ''requires'' that $S$ is non-empty. Now, assume $f$ is a surjection. Consider the indexed family of non-empty sets $\set {\map {f^{-1} } y}_{y \mathop \in T}$ where $\map {f^{-1} } y$ denotes the preimage of $y$ under $f$. Using the axiom of choice, there exists a mapping $g: T \to S$ such that $\map g y \in \set {\map {f^{-1} } y}$ for all $y \in T$. :550px That is, $\map {\paren {f \circ g} } y = y$, as desired. {{qed}}
Take the result Condition for Composite Mapping on Right: Let $A, B, C$ be sets. Let $f: B \to A$ and $g: C \to A$ be mappings. Then: :$\Img g \subseteq \Img f$ {{iff}}: :$\exists h: C \to B$ such that $h$ is a mapping and $f \circ h = g$. Let $C = A = T$, let $B = S$ and let $g = I_T$. Then the above translates into: :$\Img {I_T} \subseteq \Img f$ {{iff}}: :$\exists g: T \to S$ such that $g$ is a mapping and $f \circ g = I_T$. But we know that: :$\Img f \subseteq T = \Img {I_T}$ So by definition of set equality, the result follows. {{qed}}
The identity mapping is: :an injection, from Identity Mapping is Injection :a surjection, from Identity Mapping is Surjection and therefore a bijection. {{qed}}
The identity mapping $I_S$ is a bijection from $S$ to itself, by Identity Mapping is Bijection. {{qed}}
=== Necessary Condition === Let $f: S \to T$ be a mapping. Let $f$ be such that: :$\exists g_1: T \to S: g_1 \circ f = I_S$ :$\exists g_2: T \to S: f \circ g_2 = I_T$ 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 $f$.
Let $f$ be both an injection and a surjection. From Mapping is Injection and Surjection iff Inverse is Mapping it follows that its inverse $f^{-1}$ is a mapping. From Inverse of Inverse Relation: : $\paren {f^{-1} }^{-1} = f$ Thus the inverse of $f^{-1}$ is $f$. But then $f$, being a bijection, is by definition a mapping. So from Mapping is Injection and Surjection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection. {{qed}}
Suppose $f$ is a bijection. Because $f^{-1}$ is a bijection from Bijection iff Inverse is Bijection, it is by definition  a mapping. The result follows directly from Image of Element under Inverse Mapping. {{qed}}
Let $f: S \to T$ be a bijection. From Inverse of Bijection is Bijection, $f^{-1}$ is also a bijection. Let $x \in S$. From Inverse Element of Bijection: :$\exists y \in T: y = \map f x \implies x = \map {f^{-1} } y$ Then: {{begin-eqn}} {{eqn | l = \map {f^{-1} \circ f} x       | r = \map {f^{-1} } {\map f x}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {f^{-1} } y       | c = By Hypothesis }} {{eqn | r = x       | c = By Hypothesis }} {{eqn | r = \map {I_S} x       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} From Domain of Composite Relation and Codomain of Composite Relation, the domain and codomain of $f^{-1} \circ f$ are both $S$, and so are those of $I_S$ by definition. So all the criteria for Equality of Mappings are met and thus $f^{-1} \circ f = I_S$. Let $y \in T$. From Inverse Element of Bijection: :$\exists x \in S: x = \map {f^{-1} } y \implies y = \map f x$ Then: {{begin-eqn}} {{eqn | l = \map {f \circ f^{-1} } y       | r = \map f {\map {f^{-1} } y}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map f x       | c = By Hypothesis }} {{eqn | r = y       | c = By Hypothesis }} {{eqn | r = \map {I_T} y       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} From Domain of Composite Relation and Codomain of Composite Relation, the domain and codomain of $f \circ f^{-1}$ are both $T$, and so are those of $I_T$ by definition. So all the criteria for Equality of Mappings are met and thus $f \circ f^{-1} = I_T$. {{Qed}}
As every bijection is also by definition an injection, a composite of bijections is also a composite of injections. Every composite of injections is also an injection by Composite of Injections is Injection. As every bijection is also by definition a surjection, a composite of bijections is also a composite of surjections. Every composite of surjections is also a surjection by Composite of Surjections is Surjection. As a composite of bijections is therefore both an injection and a surjection, it is also a bijection. {{Qed}}
Let $g: X \to Y$ and $f: Y \to Z$ be bijections. Then from Bijection iff Inverse is Bijection, both $f^{-1}$ and $g^{-1}$ are bijections. From Inverse of Composite Relation we have that $g^{-1} \circ f^{-1}$ is the inverse of $f \circ g$. Then: :$\paren {f \circ g} \circ \paren {g^{-1} \circ f^{-1} } = I_Z$ :$\paren {g^{-1} \circ f^{-1} } \circ \paren {f \circ g} = I_X$ Hence the result.