nyuuzyou/wikis huggingface dataset:
Universal Closures are Semantically Equivalent

== Theorem ==
Let $\mathbf A$ be a WFF of predicate logic.
Let $\mathbf B, \mathbf B'$ be universal closures of $\mathbf A$.
Then $\mathbf B$ and $\mathbf B'$ are semantically equivalent.
== Proof ==
Let $\AA$ be a structure for predicate logic.
Let $\mathbf B$ be any universal closure of $\mathbf A$.
Then $\mathbf B$ is a sentence of the form:
:$\forall x_1: \cdots \forall x_n: \mathbf A$
By definition of the models relation:
:$\AA \models_{\mathrm{PL}} \mathbf B$ $\map {\operatorname{val}_\AA} {\mathbf B} = \T$ 
Hence, recursively applying the definition of $\map {\operatorname{val}_\AA} \cdot \sqbrk \sigma$, we see:
:$\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \O = \T$ $\forall a_1, \ldots, a_n \in A: \mathop{\map {\operatorname{val}_\AA} {\mathbf A} } \sqbrk {\dfrac{x_1} {a_1} + \ldots + \dfrac{x_n} {a_n} } = \T$
where $\dfrac{x_1} {a_1} + \ldots + \dfrac{x_n} {a_n}$ denotes the iterated extension of an assignment.
By Value of Formula under Assignment Determined by Free Variables:
$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\dfrac{x_1} {a_1} + \ldots + \dfrac{x_n} {a_n} }$ only depends on the $a_i$ for the free variables $x_i$ in $\mathbf A$.
Because we check all possible $a_i \in A$ and all free variables $x_i$ in $\mathbf A$ are quantified over in $\mathbf B$, it follows that:
:$\forall a_1, \ldots, a_n \in A: \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\dfrac{x_1} {a_1} + \ldots + \dfrac{x_n} {a_n} } = \T$
:
:$\forall a_1, \ldots, a_k \in A: \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\dfrac{x_1} {a_1} + \ldots + \dfrac{x_k} {a_k} } = \T$
where $x_1, \ldots, x_k$ are the free variables of $\mathbf A$.
But this last condition does not depend on $\mathbf B$ beyond that it be a universal closure.
Hence, for any two universal closures $\mathbf B, \mathbf B'$ of $\mathbf A$:
:$\AA \models_{\mathrm{PL}} \mathbf B$ $\AA \models_{\mathrm{PL}} \mathbf B$
The result follows by definition of semantic equivalence.
== Sources ==
* : $\text{II}.8$ Further Semantic Notions: Lemma $\text{II}.8.3$
Category:Predicate Logic
Universal Generalisation/Informal Statement

== Theorem ==
Let $\mathbf a$ be any arbitrarily selected object in the universe of discourse.
Then:
In natural language:
:Suppose $P$ is true of any arbitrarily selected $\mathbf a$ in the universe of discourse.''
:Then $P$ is true of everything in the universe of discourse.
== Proof ==
We can express $\forall x$ using its propositional expansion:
:$\map P {\mathbf X_1} \land \map P {\mathbf X_2} \land \map P {\mathbf X_3} \land \ldots$
where $\mathbf X_1, \mathbf X_2, \mathbf X_3 \ldots{}$ is the complete set of the objects in the universe of discourse.
The fact that any object we care to choose has the property in question means that they all must have this property.
The result then follows by generalising the Rule of Conjunction.
== Sources ==
* : $4$: Propositional Functions and Quantifiers: $4.2$: Proving Validity: Preliminary Quantification Rules
* : $\S \text{IV}$: The Logic of Predicates $(2): \ 2$: Universal Instantiation
Category:Predicate Logic
Universal Generalisation/Proof System

== Theorem ==
Let $\LL$ be a specific signature for the language of predicate logic.
Let $\mathscr H$ be Hilbert proof system instance 1 for predicate logic.
Let $\map {\mathbf A} x$ be a WFF of $\LL$.
Let $\FF$ be a collection of WFFs of $\LL$.
Let $c$ be an arbitrary constant symbol which is not in $\LL$.
Let $\LL'$ be the signature $\LL$ extended with the constant symbol $c$.
Suppose that we have the provable consequence (in $\LL'$):
:$\FF \vdash_{\mathscr H} \map {\mathbf A} c$
Then we may infer (in $\LL$):
:$\FF \vdash_{\mathscr H} \forall x: \map {\mathbf A} x$
== Proof ==
== Sources ==
* : $\text{II}.11$ Some Strategies for Constructing Proofs: Lemma $\text{II}.11.8$: Quantifier Rules (result statement)
Category:Predicate Logic
Category:Hilbert Proof System Instance 1 for Predicate Logic
Universal Instantiation/Informal Statement

== Theorem ==
Suppose we have a universal statement:
:$\forall x: \map P x$
where $\forall$ is the universal quantifier and $\map P x$ is a propositional function.
Then we can deduce:
:$\map P {\mathbf a}$
where $\mathbf a$ is any arbitrary object we care to choose in the universe of discourse.
In natural language:
:Suppose $P$ is true of everything in the universe of discourse.
:Let $\mathbf a$ be an element of the universe of discourse.
:Then $P$ is true of $\mathbf a$.
== Proof ==
In the language of symbolic logic:
== Sources ==
* : $4$: Propositional Functions and Quantifiers: $4.2$: Proving Validity: Preliminary Quantification Rules
* : $\S \text{IV}$: The Logic of Predicates $(2): \ 2$: Universal Instantiation
Category:Universal Instantiation
Universal Instantiation/Model

== Theorem ==
Let $\map {\mathbf A} x$ be a WFF of predicate logic.
Let $\tau$ be a term which is freely substitutable for $x$ in $\mathbf A$.
Then $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a tautology.
== Proof ==
Let $\AA$ be a structure on a set $A$, and let $\sigma$ be an assignment for $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$.
Define:
:$a_\tau := \mathop {\map {\operatorname{val}_\AA} \tau} \sqbrk \sigma$
the value of $\tau$ under $\sigma$.
From the definition of value under $\sigma$:
:$\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau} \sqbrk \sigma = \map {f^\to} {\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x} \sqbrk \sigma, \map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma}$
where $f^\to$ is the truth function of $\implies$.
We thus need to ascertain that if:
:$\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x} \sqbrk \sigma = T$
then also:
:$\map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma = T$
By definition of value under $\sigma$, the former amounts to:
:$\map {\mathrm {val}_\AA} {\map {\mathbf A} x} \sqbrk {\sigma + \paren {x / a} } = T$
for all $a \in A$.
By the Substitution Theorem for Well-Formed Formulas:
:$\map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma = \map {\mathrm {val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a_\tau} }$
Since $a_\tau \in A$, the conclusion follows, and $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a tautology.
== Sources ==
* : $\text {II}.8$ Further Semantic Notions: Corollary $\text {II.8.12}$
Category:Universal Instantiation
Universal Instantiation/Proof System

== Theorem ==
Let $\map {\mathbf A} x$ be a WFF of predicate logic.
Let $\tau$ be a term which is freely substitutable for $x$ in $\mathbf A$.
Let $\mathscr H$ be Hilbert proof system instance 1 for predicate logic.
Then:
:$\forall x: \map {\mathbf A} x \vdash_{\mathscr H} \map {\mathbf A} \tau$
is a provable consequence in $\mathscr H$.
== Proof ==
== Sources ==
* : $\text{II}.11$ Some Strategies for Constructing Proofs: Lemma $\text{II}.11.8$: Quantifier Rules (result statement)
Category:Universal Instantiation
Category:Hilbert Proof System Instance 1 for Predicate Logic
Universal Negative implies Particular Negative iff First Predicate is not Vacuous

== Theorem ==
Consider the categorical statements:
Then:
:$\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$
:
:$\exists x: \map S x$
Using the symbology of predicate logic:
:$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \neg \map P x} \implies \paren {\exists x: \map S x \land \neg \map P x} }$
== Proof ==
=== Sufficient Condition ===
Let $\exists x: \map S x$.
Let $\map {\mathbf E} {S, P}$ be true.
As $\map {\mathbf E} {S, P}$ is true, then by Modus Ponendo Ponens:
:$\neg \map P x$
From the Rule of Conjunction:
:$\map S x \land \neg \map P x$
Thus $\map {\mathbf O} {S, P}$ holds.
So by the Rule of Implication:
:$\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$
=== Necessary Condition ===
Let $\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$.
:
:$\neg \exists x: \map S x$
that is, $\map S x$ is vacuous.
From De Morgan's Laws: Denial of Existence:
:$\forall x: \neg \map S x \dashv \vdash \neg \exists x: \map S x$
it follows that $\forall x: \map S x$ is false.
From False Statement implies Every Statement:
:$\forall x: \map S x \implies \neg \map P x$
is true.
So $\map {\mathbf E} {S, P}$ holds.
Again, $\neg \exists x: \map S x$.
Then by the Rule of Conjunction:
:$\neg \paren {\exists x: \map S x \land \neg \map P x}$
That is, $\map {\mathbf O} {S, P}$ does not hold.
So $\map {\mathbf E} {S, P}$ is true and $\map {\mathbf O} {S, P}$ is false.
This contradicts $\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$ by definition of implication.
Thus $\exists x: \map S x$ must hold.
== Also defined as ==
Some sources gloss over the possibility of $\map S x$ being vacuous and merely report that the universal negative implies the particular negative.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
* : $4$: Propositional Functions and Quantifiers: $4.1$: Singular Propositions and General Propositions
Category:Categorical Statements
Universal Property for Field of Quotients

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Let $\struct {F, \oplus, \cdot}$ be a field of quotients of $D$.
Then $F$ satisfies the following universal property:
There exists a (ring) homomorphism $\iota : D \to F$ such that:
::for every field $\tilde F$ and
:and:
::for every (ring) homomorphism $\phi: D \to \tilde F$
:there exists a unique field homomorphism $\psi: F \to \tilde F$ satisfying:
:::$\psi \iota = \phi$
That is, the following diagram commutes:
:
Namely we may take:
:$\psi: a / b \mapsto \map \phi a \map \phi b^{-1}$
== Proof ==
Category:Fields of Quotients
Category:Integral Domains
Category:Universal Properties
Universal Property for Simple Field Extensions

== Theorem ==
Let $F / K$ be a field extension.
Let $\alpha \in F$ be an algebraic over $K$.
Let $\mu_\alpha$ be the minimal polynomial over $\alpha$ over $K$.
Let $\psi : K \left({\alpha}\right) \to L$ be a homomorphism.
Let $\phi = \psi \restriction_K$.
Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the Induced Homomorphism of Polynomial Forms.
Then $\psi\left({\alpha}\right)$ is a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
Conversely let $L$ be a field.
Let $\phi: K \to L$ be a homomorphism.
Let $\beta$ be a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
Then there exists a unique field homomorphism $\psi : K \left({\alpha}\right) \to L$ extending $\phi$ that sends $\alpha$ to $\beta$.
== Proof ==
Let $\psi: K \left({\alpha}\right) \to L$ be a homomorphism.
Let $\phi = \psi \big|_K$.
Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the Induced Homomorphism of Polynomial Forms.
For any $f = a_0 + \dotsb + a_n X^n \in K \left[{X}\right]$ we have:
Since $\mu_\alpha \left({\alpha}\right) = 0$ and a Ring Homomorphism Preserves Zero, the above yields:
:$\overline \phi \left({f}\right) \left({\psi \left({\alpha}\right)}\right) = 0$
as required.
Conversely let $L$ be a field.
Let $\phi: K \to L$ be a homomorphism.
Let $\beta$ be a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
By Structure of Simple Algebraic Field Extension, there exists a unique isomorphism:
:$\Delta: K \left[{\alpha}\right] \to K \left[{X}\right] / \langle \mu_\alpha \rangle$
such that:
:$\Delta \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
and:
:$\Delta \big|_K$ is the identity mapping.
By Universal Property of Polynomial Ring there exists a unique homomorphism:
:$\chi: K \left[{X}\right] \to L$
such that:
:$\chi \left({X}\right) = \beta$
and:
:$\chi \big|_K = \phi$
By Universal Property of Quotient Ring there exists a unique homomorphism:
:$\psi: K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle \to L$
such that:
:$\chi = \psi \circ \pi$
So we have the following diagram:
::$\begin{xy}\xymatrix@L+2mu@+1em {
 K \left[{X}\right] \ar[r]^*{\pi}
 \ar@{-->}[rd]_*{\exists ! \chi}
&
 \dfrac {K \left[{X}\right]} {\left\langle{\mu_\alpha}\right\rangle} \ar@{-->}[d]^*{\exists ! \psi}
 \ar[r]^*{\Delta}
&
 K \left[{\alpha}\right]
\\ 
&
 L
&
}\end{xy}$
Now we have:
:$\chi \left({X}\right) = \beta$
and:
:$\pi \left({X}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
Therefore, because:
:$\chi = \psi \circ \pi$
we have:
:$\psi \left({X + \left\langle{\mu_\alpha}\right\rangle}\right) = \beta$
Also:
:$\Delta^{-1} \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
so by the above:
: $\psi \circ \Delta^{-1}\left({ \alpha }\right) = \beta$
: $\psi \circ \Delta^{-1} \big|_K = \psi\big|_K = \phi$
: the choice of $\Delta, \psi$ is unique.
Category:Field Extensions
Category:Universal Properties
Universal Property of Abelianization of Group

== Theorem ==
Let $G$ be a group.
Let $G^{\operatorname {ab} }$ be its abelianization.
Let $\pi : G \to G^{\operatorname {ab} }$ be the quotient group epimorphism.
Let $H$ be an abelian group.
Let $f: G \to H$ be a group homomorphism.
Then there exists a unique group homomorphism $g : G^{\operatorname {ab}} \to H$ such that $g \circ \pi = f$:
:$\xymatrix {
G \ar[d]_\pi \ar[r]^{\forall f} & H\\
G^{\operatorname {ab} } \ar[ru]_{\exists ! g} }$
== Proof 1 ==
== Proof 2 ==
Category:Universal Property of Abelianization of Group
Category:Abelianizations of Groups
Category:Universal Properties
Universal Property of Direct Product of Modules

== Theorem ==
Let $R$ be a ring.
Let $N$ be an $R$-module.
Let $\family{M_i}_{i \mathop \in I}$ be a family of $R$-modules.
Let $M = \ds \prod_{i \mathop \in I} M_i$ be their direct product.
Let $\family{\psi_i}_{i \mathop \in I}$ be a family of $R$-module morphisms $N \to M_i$.
Then there exists a unique morphism:
:$\Psi: N \to M$
such that:
: $\forall i: \psi_i = \pi_i \circ \Psi$
where $\pi_i: M \to M_i$ is the $i$th canonical projection.
== Proof ==
== Also see ==
* Universal Property of Direct Sum of Modules
Category:Module Theory
Category:Universal Properties
Category:Direct Products
Universal Property of Direct Sum of Modules

== Theorem ==
Let $R$ be a ring.
Let $N$ be an $R$-module.
Let $\family {M_i}_{i \mathop \in I}$ be a family of $R$-modules.
Let $M = \ds \bigoplus_{i \mathop \in I} M_i$ be their direct sum.
Let $\family {\psi_i}_{i \mathop \in I}$ be a family of $R$-module morphisms $M_i \to N$.
Then there exists a unique morphism:
:$\Psi: M \to N$
such that:
:$\forall i: \psi_i = \Psi \circ \iota_i$
where $\iota_i: M_i \to M$ is the $i$th canonical injection.
== Proof ==
== Also see ==
* Universal Property of Direct Product of Modules
Category:Module Theory
Category:Direct Sums
Category:Universal Properties
Universal Property of Field of Rational Fractions

== Theorem ==
Let $R$ be an integral domain.
Let $\struct {\map R x, \iota, x}$ be the field of rational fractions over $R$.
Let $\struct {K, f, a}$ be an ordered triple, where:
:$K$ is a field
:$f : R \to K$ is a unital ring homomorphism
:$a$ is a transcendental element of $K$.
Then there exists a unique unital ring homomorphism $\bar f : \map R x \to K$ such that $\bar f \circ \iota = f$ and $\map {\bar f} x = a$:
:$\xymatrix{
R \ar[d]^\iota \ar[r]^{\forall f} & K\\
\map R x \ar[ru]_{\exists ! \bar f} }$
== Proof ==
Use Universal Property of Polynomial Ring and Universal Poperty of Field of Fractions.
Category:Polynomial Theory
Category:Universal Properties
Universal Property of Free Abelian Group on Set

== Theorem ==
Let $S$ be a set.
Let $\struct {\Z^{\paren S}, \iota}$ be the free abelian group on $S$.
Let $\struct {G, +}$ be an abelian group.
Let $f: S \to G$ be a mapping.
Then there exists a unique group homomorphism $g : \Z^{\paren S} \to G$ with $g \circ \iota = f$:
:$\xymatrix{S \ar[d]_\iota \ar[r]^{\forall f} & G\\ \Z^{\paren S} \ar@{-->}[ru]_{\exists ! g} & }$
== Proof ==
For $x \in S$, the characteristic map $\phi_x$ of $x$ is defined as:
:$\map {\phi_x} s = \begin{cases} 1 & \quad \text{ if } s = x \\ 0 & \quad \text{ if } s \ne x \end{cases}$
The maps $\set {\phi_x \ : \ x \in S}$ form a basis for $\Z^{\paren S}$.
If $\phi \in \Z^{\paren S}$, then it can be expressed uniquely as a finite sum 
:$\ds \phi = \sum_i n_i \phi_{x_i}$
where $\phi$ maps each $x_i$ to $n_i \in \Z$, and maps each element of $S$ not present in this sum to $0$.
$\iota$ maps $x \in S$ to $\phi_x \in \Z^{\paren S}$.
So, in order to satisfy $g \circ \iota = f$, we must have $\map g {\phi_x} = \map f x$ for every $x \in S$. 
Since $g$ has to be a homomorphism, it must satisfy $\map g {\phi + \phi'} = \map g \phi + \map g {\phi'}$.
So, for $\phi \in \Z^{\paren S}$:
:$\ds \map g \phi = \map g {\sum_i n_i \phi_{x_i} } = \sum_i \map g {n_i \phi_{x_i} } = \sum_i n_i \map g {\phi_{x_i} } = \sum_i n_i \map f {x_i}$
Therefore, the value of $\map g \phi$ is unique, for every $\phi \in \Z^{\paren S}$.
So, the homomorphism $g$ that satisfies $g \circ \iota = f$ is unique.
== Also see ==
* Universal Property of Free Module on Set
Category:Universal Properties
Category:Abelian Groups
Universal Property of Free Module on Set

== Theorem ==
Let $R$ be a ring with unity.
Let $R^{\paren I}$ be the free $R$-module on $I$.
Let $M$ be an $R$-module.
Let $\family {m_i}_{i \mathop \in I}$ be a family of elements of $M$.
Then there exists a unique $R$-module morphism:
:$\Psi: R^{\paren I} \to M$
that sends the $i$th canonical basis element to $m_i$, for all $i \in I$.
Moreover:
:$\ds \map \Psi {\family {r_i}_{i \mathop \in I} } = \sum_{i \mathop \in I} r_i m_i$
== Proof ==
=== Existence ===
By Morphism from Ring with Unity to Module, for all $i$ there exists a morphism:
:$\psi_i: R \to M$
with $\map {\psi_i} 1 = m_i$.
By Universal Property of Direct Sum of Modules, there exists a morphism:
:$\Psi: R^{\paren I} \to M$
such that $\Psi \circ \iota_i = \psi_i$ for all $i$.
Thus:
:$\forall i \in I: \map \Psi {e_i} = \map \Psi {\map {\iota_i} 1} = \map {\psi_i} 1 = m_i$
We have:
:$\ds \family {r_i}_{i \mathop \in I} = \sum_{i \mathop \in I} r_i e_i$
so the expression for $\Psi$ follows by linearity.
=== Uniqueness ===
Let $\Psi$ be such a morphism.
Then $\Psi \circ \iota_i$ sends $1$ to $m_i$.
By Morphism from Ring with Unity to Module:
:$\Psi \circ \iota_i = \psi_i$
with $\psi_i$ as above.
By Universal Property of Direct Sum of Modules, $\Psi$ is determined by $\Psi \circ \iota_i$.
Thus $\Psi$ is unique.
Category:Module Theory
Category:Direct Products
Category:Universal Properties
Universal Property of Free Modules

== Theorem ==
Let $R$ be a ring.
Let $M$ be a free $R$-module with basis $\{e_i\mid i\in I\}$.
Let $N$ be an $R$-module.
Let $\{n_i\mid i\in I\}$ be a family of elements of $N$.
Then there exists a unique $R$-module homomorphism that maps $e_i$ to $n_i$ for all $i\in I$.
== Proof ==
Combine Free Module is Isomorphic to Free Module Indexed by Set and Universal Property of Free Module on Set.
Category:Free Modules
Category:Module Theory
Category:Universal Properties
Universal Property of Group Ring

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be a group.
Let $R \sqbrk G$ be the corresponding group ring.
Let $S$ be a commutative ring with unity.
Let $\phi: R \to S$ be a ring homomorphism.
Let $\beta : G \to R^\times$ be a group homomorphism, where $R^\times$ is the multiplicative group of $R$.
Then there exists a unique ring homomorphism from $R \sqbrk G$ to $S$ which extends $\phi$ and $\beta$.
== Proof ==
Category:Universal Properties
Category:Group Rings
Universal Property of Polynomial Ring

== Theorem ==
Let $R$ be a commutative ring with unity.
The different definitions of a polynomial ring $\struct {\map R x, \iota, x}$ on $R$ satisfy the universal property of a polynomial ring:
=== Construction using Free Monoid on Set ===
== Also see ==
* Equivalence of Definitions of Polynomial Ring
Category:Polynomial Theory
Category:Universal Properties
Universal Property of Polynomial Ring/Free Monoid on Set

== Theorem ==
Let $R, S$ be commutative and unitary rings.
Let $\family {s_j}_{j \mathop \in J}$ be an indexed family of elements of $S$.
Let $\psi: R \to S$ be a ring homomorphism.
Let $R \sqbrk {\set {X_j: j \in J} }$ be a polynomial ring.
Then there exists a unique evaluation homomorphism $\phi: R \sqbrk {\set {X_j: j \in J} } \to S$ at $\family {s_j}_{j \mathop \in J}$ extending $\psi$.
== Proof ==
Let $Z$ be the set of all multiindices indexed by $J$.
Let $k_j$ be the $j$th component of a multiindex $k$.
Let $\ds f = \sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j}$ be a polynomial over $R$.
Define:
:$\ds \map \phi f = \sum_{k \mathop \in Z} \map \psi {a_k} \prod_{j \mathop \in J} s_j^{k_j}$
It is clear that $\phi$ extends $\psi$.
If $\ds g = \sum_{k \mathop \in Z} b_k \prod_{j \mathop \in J} X_j^{k_j}$, then:
 
Therefore $\phi$ preserves addition.
Also:
This shows that $\phi$ is a homomorphism.
Now suppose that $\phi'$ is another such homomorphism.
For each $j \in J$, $\phi'$ must satisfy $\map {\phi'} {X_j} = s_j$ and $\map {\phi'} r = \map \psi r$ for all $r \in R$.
In addition $\phi'$ must be a homomorphism, so we compute:
and therefore $\phi' = \phi$.
This concludes the proof.
== Remarks ==
* The requirement that the rings be commutative is vital. A fundamental difference for polynomials over non-commutative rings is additional difficulty identifying polynomial forms and functions using this method.
== Also see ==
* Equivalence of Definitions of Polynomial Ring
* Universal Property of Polynomial Algebra
Category:Polynomial Theory
Category:Universal Properties
Universal Property of Quotient Group

== Theorem ==
Let $G$ and $H$ be groups.
Let $N \trianglelefteq G$ be an normal subgroup.
Let $\pi: G \to G / N$ be the quotient epimorphism.
Let $f: G \to H$ be a group homomorphism with $N \subseteq \ker f$.
Then there exists a unique group homomorphism $\overline f: G / N \to H$ such that $f = \overline f \circ \pi$.
$\xymatrix{
G \ar[d]^\pi \ar[r]^{\forall f} & H \\
G/N \ar[ru]_{\exists ! \bar f} }$
== Proof ==
=== Existence ===
Let $\sim$ denote (left) congruence modulo $N$.
From Congruence Modulo Subgroup is Equivalence Relation, $\sim$ is an equivalence relation on $X$.
For all $g \in G$, let $\eqclass g \sim$ denote the equivalence class of $g$ under $\sim$.
Note that Group Homomorphism is Invariant under Congruence Modulo Kernel.
In particular, since $N \subseteq \ker f$, we have that $f$ is $\sim$-invariant.
Hence, by Universal Property of Quotient Set, there exists a unique such mapping $\overline f: G / \sim{} \to H$, which is given by:
:$\forall g \in G: \map {\overline f} {\eqclass g \sim} = \map f g$
From Universal Property of Quotient Set, $f$ is well-defined.
By definition of (left) coset space, $G / N = G / \sim$. 
By Left Cosets are Equal iff Product with Inverse in Subgroup:
:for all $g \in G$ the equivalence class $\eqclass g \sim$ is the (left) coset of $N$ containing $g$, which is $g N$.
Thus $\overline f$ is given by:
:$\forall g \in G: \map {\overline f} {g N} = \map f g$
We have that $\overline f$ is a group homomorphism.
Indeed, for all $x, y \in G$ we have:
Thus we have shown the existence of such a group homomorphism.
=== Uniqueness ===
By Universal Property of Quotient Set, there exists a unique such mapping $\overline f$.
A fortiori, there exists at most one such group homomorphism.
== Also see ==
* First Isomorphism Theorem for Groups
* Universal Property of Quotient of Topological Group
Category:Quotient Groups
Category:Universal Properties