hcju/proofwikiqa huggingface dataset:
$\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ is a specific example of Inverse of Composite Relation. As $f$ and $g$ are bijections then by Bijection iff Inverse is Bijection, so are both $f^{-1}$ and $g^{-1}$. By Composite of Bijections is Bijection, it follows that $f^{-1} \circ g^{-1}$ is a bijection. {{qed}}
Let $f: X \to Y$ and $g: Y \to Z$ be bijections. Then: {{begin-eqn}} {{eqn | l = \paren {g \circ f} \circ \paren {f^{-1} \circ g^{-1} }       | r = g \circ \paren {\paren {f \circ f^{-1} } \circ g^{-1} }       | c = Composition of Mappings is Associative }} {{eqn | r = g \circ \paren {I_Y \circ g^{-1} }       | c = Composite of Bijection with Inverse is Identity Mapping }} {{eqn | r = g \circ g^{-1}       | c = Identity Mapping is Left Identity }} {{eqn | r = I_Z       | c = Composite of Bijection with Inverse is Identity Mapping }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {f^{-1} \circ g^{-1} } \circ \paren {g \circ f}       | r = \paren {f^{-1} \circ \paren {g^{-1} \circ g} } \circ f       | c = Composition of Mappings is Associative }} {{eqn | r = \paren {f^{-1} \circ I_Y} \circ f       | c = Composite of Bijection with Inverse is Identity Mapping }} {{eqn | r = f^{-1} \circ f       | c = Identity Mapping is Right Identity }} {{eqn | r = I_X       | c = Composite of Bijection with Inverse is Identity Mapping }} {{end-eqn}} Hence the result.
This follows from the fact that a permutation is a bijection.  The domain and codomain are coincident. The result follows from Composite of Bijections is Bijection. {{Qed}}
Let $f: S \to S$ is a permutation of $S$. By definition, a permutation is a bijection such that the domain and codomain are the same set. From Bijection iff Inverse is Bijection, it follows $f^{-1}$ is a bijection. From the definition of inverse relation, the domain of a relation is the codomain of its inverse and vice versa. Thus the domain and codomain of $f^{-1}$ are both $S$ and it follows that $f^{-1}$ is a permutation. {{Qed}}
Let $f: S \to T$ be an injection. Then from Injection iff Left Inverse there exists at least one left inverse $g: T \to S$ of $f$ such that $g \circ f = I_S$. $I_S$ is a surjection. Thus $g \circ f$ is a surjection. By Surjection if Composite is Surjection it follows that $g$ is also a surjection. {{Qed}}
By the definition of right inverse: :$f \circ g = I_T$ where $I_T$ is the identity mapping on $T$. By Identity Mapping is Injection, $I_T$ is an injection. By Injection if Composite is Injection, it follows that $g$ is an injection. {{qed}}
For two sets to be equivalent, there needs to exist a bijection between them. In the following, let $\phi$, $\phi_1$, $\phi_2$ etc. be understood to be bijections. === Reflexive === From Identity Mapping is Bijection, the identity mapping $I_S: S \to S$ is the required bijection. Thus there exists a bijection from $S$ to itself and $S$ is therefore equivalent to itself. Therefore set equivalence is reflexive. {{qed|lemma}} === Symmetric === {{begin-eqn}} {{eqn | o =        | r = S \sim T       | c =  }} {{eqn | o = \leadsto       | r = \exists \phi: S \to T       | c = {{Defof|Set Equivalence}} }} {{eqn | o = \leadsto       | r = \exists \phi^{-1}: T \to S       | c = Bijection iff Inverse is Bijection }} {{eqn | o = \leadsto       | r = T \sim S       | c = {{Defof|Set Equivalence}}: $\phi^{-1}$ is also a bijection }} {{end-eqn}} Therefore set equivalence is symmetric. {{qed|lemma}} === Transitive === {{begin-eqn}} {{eqn | o =        | r = S_1 \sim S_2 \land S_2 \sim S_3       | c =  }} {{eqn | o = \leadsto       | r = \exists \phi_1: S_1 \to S_2 \land \exists \phi_2: S_2 \to S_3       | c = {{Defof|Set Equivalence}} }} {{eqn | o = \leadsto       | r = \exists \phi_2 \circ \phi_1: S_1 \to S_3       | c = Composite of Bijections is Bijection: $\phi_2 \circ \phi_1$ is a bijection }} {{eqn | o = \leadsto       | r = S_1 \sim S_3       | c = {{Defof|Set Equivalence}} }} {{end-eqn}} Therefore set equivalence is transitive. {{qed}} {{expand|See talk}}
As $S \sim T$, there exists a bijection $f: S \to T$. Let $g: \left({S \setminus \left\{{a}\right\}}\right) \to \left({T \setminus \left\{{b}\right\}}\right)$ be the mapping defined as follows: $\forall x \in S \setminus \left\{{a}\right\}: g \left({x}\right) = \begin{cases} f \left({x}\right): f \left({x}\right) \ne b \\ f \left({a}\right): f \left({x}\right) = b \end{cases}$ First it is shown that $g$ is an injection. Let $x, y \in S \setminus \left\{{a}\right\}$ and $x \ne y$. The following cases apply: $(1): \quad f \left({x}\right) \ne b$ and $f \left({y}\right) \ne b$ By definition of $g$: :$f \left({x}\right) = g \left({x}\right)$ and: :$f \left({y}\right) = g \left({y}\right)$ It follows from the injectivity of $f$ that: :$g \left({x}\right) \ne g \left({y}\right)$ $(2): \quad f \left({x}\right) = b$ By definition of $g$: :$g \left({x}\right) = f \left({a}\right)$ We have that: :$y \in S \setminus \left\{{a}\right\}$ Therefore by definition of set difference: :$y \ne a$ So by the injectivity of $f$: :$f \left({y}\right) \ne f \left({a}\right) = g \left({x}\right)$ Also by the injectivity of $f$: :$f \left({y}\right) \ne b$ Thus, by the definition of $g$: :$g \left({y}\right) = f \left({y}\right) \ne f \left({a}\right) = g \left({x}\right)$ $(2): \quad f \left({y}\right) = b$ The proof is the same as that of case $2$, with the roles of $x$ and $y$ reversed. {{qed|lemma}} Next it is shown that $g$ is a surjection. Let $y \in T \setminus \left\{{b}\right\}$. By definition of surjectivity of $f$: :$\exists x \in S: f \left({x}\right) = y$ It is to be shown that: :$\exists x' \in S, x \ne a: f \left({x'}\right) = y$ $(1): \quad f \left({a}\right) = b$ By hypothesis: :$f \left({x}\right) = y \ne b$ Then by the rule of transposition: :$x \ne a$ so let: :$x' = x$ $(2): \quad f \left({a}\right) \ne b$ By the rule of transposition: :$f^{-1} \left({b}\right) \ne a$ where $f^{-1} \left({b}\right)$ denotes the preimage of $b$ under $f$. :Case $(2.1): \quad f \left({a}\right) = y$ By the definition of $g$: :$g \left({f^{-1} \left({b}\right)}\right) = y$ so let: :$x' = f^{-1} \left({b}\right)$ :Case $(2.2): \quad f \left({a}\right) \ne y$ By hypothesis: :$f \left({x}\right) = y \ne f \left({a}\right)$ Then by the rule of transposition: :$x \ne a$ so let: :$x' = x$ Thus: :$\forall y \in T \setminus \left\{{b}\right\}: \exists x \in S \setminus \left\{{a}\right\}: g \left({x}\right) = y$ and so $g$ is a surjection. {{qed|lemma}} Thus $g$ is both an injection and a surjection, and is by definition a bijection. {{qed}}
As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation: :$\RR \sqbrk {A \cup B} = \RR \sqbrk A \cup \RR \sqbrk B$ {{qed}}
First we have: {{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = A \cup B       | c = Set is Subset of Union }} {{eqn | ll= \leadsto       | l = f \sqbrk A       | o = \subseteq       | r = f \sqbrk {A \cup B}       | c = Image of Subset under Mapping is Subset of Image }} {{end-eqn}} {{begin-eqn}} {{eqn | l = B       | o = \subseteq       | r = A \cup B       | c = Set is Subset of Union }} {{eqn | ll= \leadsto       | l = f \sqbrk B       | o = \subseteq       | r = f \sqbrk {A \cup B}       | c = Image of Subset under Mapping is Subset of Image }} {{end-eqn}} {{begin-eqn}} {{eqn | ll= \leadsto       | l = f \sqbrk A \cup f \sqbrk B       | o = \subseteq       | r = f \sqbrk {A \cup B}       | c = Union is Smallest Superset }} {{end-eqn}} {{Qed|lemma}} Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = f \sqbrk {A \cup B}       | c =  }} {{eqn | ll= \leadsto       | l = \exists x \in A \cup B: y       | r = \map f x       | c = {{Defof|Image of Subset under Relation}} }} {{eqn | ll= \leadsto       | l = \exists x: x \in A \lor x \in B: y       | r = \map f x       | c = {{Defof|Set Union}} }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = f \sqbrk A \lor y \in f \sqbrk B       | c = {{Defof|Image of Subset under Relation}} }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = f \sqbrk A \cup f \sqbrk B       | c = {{Defof|Set Union}} }} {{eqn | ll= \leadsto       | l = f \sqbrk {A \cup B}       | o = \subseteq       | r = f \sqbrk A \cup f \sqbrk B       | c = {{Defof|Subset}} }} {{end-eqn}} {{Qed|lemma}} Thus we have: {{begin-eqn}} {{eqn | l = f \sqbrk A \cup f \sqbrk B       | o = \subseteq       | r = f \sqbrk {A \cup B}       | c =  }} {{eqn | l = f \sqbrk {A \cup B}       | o = \subseteq       | r = f \sqbrk A \cup f \sqbrk B       | c =  }} {{eqn | ll= \leadsto       | l = f \sqbrk A \cup f \sqbrk B       | r = f \sqbrk {A \cup B}       | c = {{Defof|Set Equality}} }} {{end-eqn}} {{qed}}
As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation: :$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ {{qed}}
{{begin-eqn}} {{eqn | l = S_1 \cap S_2       | o = \subseteq       | r = S_1       | c = Intersection is Subset }} {{eqn | ll= \leadsto       | l = f \sqbrk {S_1 \cap S_2}       | o = \subseteq       | r = f \sqbrk {S_1}       | c = Image of Subset under Mapping is Subset of Image }} {{end-eqn}} {{begin-eqn}} {{eqn | l = S_1 \cap S_2       | o = \subseteq       | r = S_2       | c = Intersection is Subset }} {{eqn | ll= \leadsto       | l = f \sqbrk {S_1 \cap S_2}       | o = \subseteq       | r = f \sqbrk {S_2}       | c = Image of Subset under Mapping is Subset of Image }} {{end-eqn}} {{begin-eqn}} {{eqn | ll= \leadsto       | l = f \sqbrk {S_1 \cap S_2}       | o = \subseteq       | r = f \sqbrk {S_1} \cap f \sqbrk {S_2}       | c = Intersection is Largest Subset }} {{end-eqn}} {{qed}}
An injection is a type of one-to-one relation, and therefore also a one-to-many relation. Therefore Image of Intersection under One-to-Many Relation applies: :$\forall A, B \subseteq S: \mathcal R \left[{A \cap B}\right] = \mathcal R \left[{A}\right] \cap \mathcal R \left[{B}\right]$ {{iff}} $\mathcal R$ is a one-to-many relation. We have that $f$ is a mapping and therefore a many-to-one relation. So $f$ is a one-to-many relation {{iff}} $f$ is also an injection. It follows that: :$\forall A, B \subseteq S: f \left[{A \cap B}\right] = f \left[{A}\right] \cap f \left[{B}\right]$ {{iff}} $f$ is an injection. {{qed}}
From Image of Intersection under Mapping: :$f \sqbrk {A \cap B} \subseteq f \sqbrk A \cap f \sqbrk B$ which holds for all mappings. It remains to be shown that: :$f \sqbrk A \cap f \sqbrk B \subseteq f \sqbrk {A \cap B}$ {{iff}} $f$ is an injection. === Sufficient Condition === Suppose that: :$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$ If $S$ is singleton, then $f$ is a fortiori an injection from Mapping from Singleton is Injection. So, assume $S$ is not singleton. {{AimForCont}} $f$ is specifically ''not'' an injection. Then: :$\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$ and of course $\set x \subseteq S, \set y \subseteq S$. So: :$z \in f \sqbrk {\set x}$ :$z \in f \sqbrk {\set y}$ and so by definition of intersection: :$z \in f \sqbrk {\set x} \cap f \sqbrk {\set y}$ But: :$\set x \cap \set y = \O$ Thus from the corollary to Image of Empty Set is Empty Set: :$f \sqbrk {\set x \cap \set y} = \O$ and so: :$f \sqbrk {\set x \cap \set y} \ne f \sqbrk {\set x} \cap f \sqbrk {\set y}$ But by hypothesis: :$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$ From this contradiction it follows that $f$ is an injection. {{qed|lemma}} === Necessary Condition === Let $f$ be an injection. It is necessary to show: :$f \sqbrk {S_1} \cap f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \cap S_2}$ Let $t \in f \sqbrk {S_1} \cap f \sqbrk {S_2}$. Then: {{begin-eqn}} {{eqn | l = t       | o = \in       | r = f \sqbrk {S_1} \cap f \sqbrk {S_2}       | c = }} {{eqn | lll=\leadsto       | l = t       | o = \in       | r = f \sqbrk {S_1}       | c = {{Defof|Set Intersection}} }} {{eqn | lo= \land       | l = t       | o = \in       | r = f \sqbrk {S_2}       | c = }} {{eqn | lll=\leadsto       | ll= \exists s_1 \in S_1, s_2 \in S_2:       | l = t       | r = \map f {s_1}       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | lo= \land       | l = t       | r = \map f {s_2}       | c =  }} {{eqn | ll= \leadsto       | l = \map f {s_1}       | r = \map f {s_2}       | c =  }} {{eqn | ll= \leadsto       | l = s_1       | r = s_2       | c = {{Defof|Injection}} }} {{eqn | ll= \leadsto       | l = s_1 = s_2       | o = \in       | r = S_1       | c =  }} {{eqn | lo= \land       | l = s_1 = s_2       | o = \in       | r = S_2       | c =  }} {{eqn | ll=\leadsto       | l = s_1 = s_2       | o = \in       | r = S_1 \cap S_2       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = \map f {s_1} = \map f {s_2}       | o = \in       | r = f \sqbrk {S_1 \cap S_2}       | c = Image of Element is Subset }} {{eqn | ll= \leadsto       | l = f \sqbrk {S_1} \cap f \sqbrk {S_2}       | o = \subseteq       | r = f \sqbrk {S_1 \cap S_2}       | c = {{Defof|Subset}} }} {{end-eqn}} So if $f$ is an injection, it follows that: :$f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$ {{qed|lemma}} Putting the results together, we see that: :$f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$ {{iff}} $f$ is an injection. {{qed}}
=== Sufficient Condition === Suppose that: :$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ If $S$ is singleton, the result follows immediately as $\RR$ would ''have'' to be one-to-many. So, assume $S$ is not singleton. Suppose $\RR$ is specifically ''not'' one-to-many. So: :$\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$ and of course $\set x \subseteq S, \set y \subseteq S$. So: :$z \in \RR \sqbrk {\set x}$ :$z \in \RR \sqbrk {\set y}$ and so by definition of intersection: :$z \in \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$ But: :$\set x \cap \set y = \O$ Thus from Image of Empty Set is Empty Set: :$\RR \sqbrk {\set x \cap \set y} = \O$ and so: :$\RR \sqbrk {\set x \cap \set y} \ne \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$ {{qed|lemma}} === Necessary Condition === Let $\RR$ be one-to-many. From Image of Intersection under Relation, we already have: :$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ So we just need to show: :$\RR \sqbrk {S_1} \cap \RR \sqbrk {S_2} \subseteq \RR \sqbrk {S_1 \cap S_2}$ Let $t \in \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$. Then: {{begin-eqn}} {{eqn | l = t       | o = \in       | r = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}       | c =  }} {{eqn | ll= \leadsto       | l = t       | o = \in       | r = \RR \sqbrk {S_1} \land t \in \RR \sqbrk {S_2}       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = \exists s_1 \in S_1: \tuple {s_1, t}       | o = \in       | r = \RR \land \exists s_2 \in S_2: \tuple {s_2, t} \in \RR       | c = {{Defof|Relation}} }} {{eqn | ll= \leadsto       | l = s_1       | r = s_2       | c = {{Defof|One-to-Many Relation}} }} {{eqn | ll= \leadsto       | l = s_1 = s_2       | o = \in       | r = S_1 \land s_1 = s_2 \in S_2       | c =  }} {{eqn | ll= \leadsto       | l = s_1 = s_2       | o = \in       | r = S_1 \cap S_2       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = \map \RR {s_1} = \map \RR {s_2}       | o = \in       | r = \RR \sqbrk {S_1 \cap S_2}       | c = Image of Element is Subset }} {{eqn | ll= \leadsto       | l = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}       | o = \subseteq       | r = \RR \sqbrk {S_1 \cap S_2}       | c = {{Defof|Subset}} }} {{end-eqn}} So if $\RR$ is one-to-many, it follows that: :$\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ {{qed|lemma}} Putting the results together, we see that: :$\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ {{iff}} $\RR$ is one-to-many. {{qed}} Category:Relation Theory Category:Set Intersection sotqrohpes80rjpomanui4ipx1n4v3s
=== Sufficient Condition === First, to show that $(1)$ holds if $\mathcal R$ is one-to-many. From Image of Set Difference under Relation, we already have: :$\mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right] \subseteq \mathcal R \left[{A \setminus B}\right]$ So we just need to show: :$\mathcal R \left[{A \setminus B}\right] \subseteq \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right]$ Let $t \notin \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right]$. Then $t \notin \mathcal R \left[{A}\right] \lor t \in \mathcal R \left[{B}\right]$ by De Morgan's Laws. {{explain|Go into more detail}} Suppose $t \notin \mathcal R \left[{A}\right]$. Then by definition of a relation: : $\neg \exists s \in A: \left({s, t}\right) \in \mathcal R$ By Image of Subset is Subset of Image: : $\mathcal R \left[{A \setminus B}\right] \subseteq \mathcal R \left[{A}\right]$ Thus, by definition of subset and Rule of Transposition: : $t \notin \mathcal R \left[{A}\right] \implies t \notin \mathcal R \left[{A \setminus B}\right]$ Now suppose $t \in \mathcal R \left[{B}\right]$. Then: : $\exists s \in B: \left({s, t}\right) \in \mathcal R$ Because $\mathcal R$ is one-to-many: : $\forall x \in S: \left({x, t}\right) \in \mathcal R \implies x = s$ and thus: : $x \in B$ Thus: : $x \notin A \setminus B$ and hence: : $t \notin \mathcal R \left[{A \setminus B}\right]$ So by Proof by Cases: : $t \notin \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right] \implies t \notin \mathcal R \left[{A \setminus B}\right]$ The result follows from Set Complement inverts Subsets: :$S \subseteq T \iff \complement \left({T}\right) \subseteq \complement \left({S}\right)$ {{qed|lemma}} === Necessary Condition === Now for the converse: If $(1)$ holds, it is to be shown that $\mathcal R$ is one-to-many. Let $s, t \in S$ be distinct. That is, $s \ne t$. Then in particular: : $\left\{{s}\right\} \setminus \left\{{t}\right\} = \left\{{s}\right\}$ Applying $(1)$ to these two sets, it follows that: :$\mathcal R \left[{\left\{{s}\right\}}\right] \setminus \mathcal R \left[{\left\{{t}\right\}}\right] = \mathcal R \left[{\left\{{s}\right\}}\right]$ By Set Difference with Disjoint Set, this implies that: :$\mathcal R \left[{\left\{{s}\right\}}\right] \cap \mathcal R \left[{\left\{{t}\right\}}\right] = \varnothing$ It follows that every element of $T$ can be related to at most one element of $S$. That is, $\mathcal R$ is one-to-many. {{qed}} Category:Set Difference Category:Relation Theory i0devwmi7g85kpinon19sptljw9f9jj
As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: :$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$ {{qed}}
As $f$, being a mapping, is also a many-to-one relation, it follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation. Thus Image of Intersection under One-to-Many Relation applies: :$\mathcal R \sqbrk {T_1 \cap T_2} = \mathcal R \sqbrk {T_1} \cap \mathcal R \sqbrk {T_2}$ where here $\mathcal R = f^{-1}$. {{Qed}}
From Inverse of Mapping is One-to-Many Relation, we have that $f^{-1}: T \to S$ is one-to-many. Thus we can apply One-to-Many Image of Set Difference: :$\mathcal R \sqbrk {T_1 \setminus T_2} = \mathcal R \sqbrk {T_1} \setminus \mathcal R \sqbrk {T_2}$ where in this context $\mathcal R = f^{-1}$. {{qed}}
Let $S$ and $T$ be sets such that neither is empty. Let $\pr_1: S \times T \to S$ be the first projection on $S \times T$. Then by definition of first projection: :$\forall x \in S: \exists \tuple {x, t} \in S \times T: \map {\pr_1} {x, t} = x$ Similarly, let $\pr_2: S \times T \to T$ be the second projection on $S \times T$. Then by definition of second projection: :$\forall x \in T: \exists \tuple {s, x} \in S \times T: \map {\pr_2} {s, x} = x$ Hence the result. {{qed}}
It can be observed that this is an example of a Linear Combination of Derivatives with $\lambda = \mu = 1$. {{qed}}
From Equivalence Class is not Empty, we have that: :$\forall \eqclass x \RR \in S / \RR: \exists x \in S: x \in \eqclass x \RR$ and the result follows. {{qed}}
The diagonal relation is defined as: :$\Delta_S = \set {\tuple {x, x}: x \in S} \subseteq S \times S$ From the fact that $q_{\Delta_S}$ is a quotient mapping, we know that it is a surjection. It relates each $x \in S$ to the singleton $\set x$. Thus: :$\set x = \eqclass x {\Delta_S} \subseteq S$ So $\map {q_{\Delta_S} } x = \set x$, and it follows that: :$\eqclass x {\Delta_S} = \eqclass y {\Delta_S} \implies x = y$ Thus $q_{\Delta_S}$ is injective, and therefore by definition a bijection. {{Qed}}
Define $\mathcal R_3$ to be the equivalence induced by: :$x \mapsto \eqclass {\eqclass x {\mathcal R_1} } {\mathcal R_2}$ By definition of $\mathcal R_3$: :$\eqclass {\eqclass x {\mathcal R_1} } {\mathcal R_2} = \eqclass {\eqclass y {\mathcal R_1} } {\mathcal R_2} \implies \eqclass x {\mathcal R_3} = \eqclass y {\mathcal R_3}$ Therefore, $\phi$ is well-defined. Again by definition of $\mathcal R_3$:  :$\eqclass x {\mathcal R_3} = \eqclass y {\mathcal R_3} \implies \eqclass {\eqclass x {\mathcal R_1} } {\mathcal R_2} = \eqclass {\eqclass y {\mathcal R_1} } {\mathcal R_2}$ which means $\phi$ is an injection. Lastly, note that every element of $S / \mathcal R_3$ is of the form $\eqclass x {\mathcal R_3}$ for some $x \in S$. It is now immediate from the definition of $\phi$ that it is surjective. Hence $\phi$ is a bijection, as desired. {{qed}}
By Relation Induced by Mapping is Equivalence Relation, we have that $\RR_f$ is an equivalence relation. To determine whether $r$ is well-defined, we have to determine whether $r: S / \RR_f \to \Img f$ actually defines a mapping at all. Consider a typical element $\eqclass x {\RR_f}$ of $S / \RR_f$. Suppose we were to choose another name for the class $\eqclass x {\RR_f}$. Assume that $\eqclass x {\RR_f}$ is not a singleton. For example, let us choose $y \in \eqclass x {\RR_f}, y \ne x$ such that:  :$\eqclass x {\RR_f} = \eqclass y {\RR_f}$ then: :$\map r {\eqclass x {\RR_f} } = \map r {\eqclass y {\RR_f} }$ Hence from the definition, we have: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \eqclass x {\RR_f}       | c =  }} {{eqn | ll= \leadsto       | l = \map f y       | r = \map f x       | c =  }} {{eqn | r = \map r {\eqclass x {\RR_f} }       | c =  }} {{eqn | r = \map r {\eqclass y {\RR_f} }       | c =  }} {{end-eqn}} Thus $r$ is well-defined. {{qed}}
From Condition for Mapping from Quotient Set to be Well-Defined: :there exists a mapping $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$ {{iff}}: :$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$ But by definition of the equivalence induced by the mapping $f$: :$\forall x, y \in S: \tuple {x, y} \in \RR_f \implies \map f x = \map f y$ The result follows directly. {{qed}}
From Renaming Mapping is Well-Defined, $r: S / \RR_f \to \Img f$ is a well-defined mapping. By definition, $\RR_f$ is the equivalence relation induced by the mapping $f$. Hence by definition: :$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$ From Condition for Mapping from Quotient Set to be Injection, this is precisely the condition required for $r$ to be an injection. Next it is noted that the codomain of $r$ is $\Img f$. Then from Restriction of Mapping to Image is Surjection, we have that $f_{\restriction \Img f}$ is a surjection. Hence from Condition for Mapping from Quotient Set to be Surjection it follows that $r$ is also a surjection. Thus $r$ is shown to be both an injection and a surjection, and so by definition is a bijection. {{qed}}
Let $y = \map f x$. Then by the definition of induced equivalence: :$x \in \eqclass x {\RR_f}$ By the definition of the inverse of a mapping: :$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$ Thus: :$\eqclass x {\RR_f} = \set {s \in \Dom f: \map f s = \map f x}$ By definition: :$\map {f^{-1} } y = \eqclass x {\RR_f}$ Hence the result. {{qed}}