nyuuzyou/wikis huggingface dataset:
Composite of Automorphisms is Automorphism

== Theorem ==
Let $\struct {S, \odot_1, \odot_2, \ldots, \odot_n}$ be an algebraic structure.
Let:
:$\phi: \struct {S, \odot_1, \odot_2, \ldots, \odot_n} \to \struct {S, \odot_1, \odot_2, \ldots, \odot_n}$
:$\psi: \struct {S, \odot_1, \odot_2, \ldots, \odot_n} \to \struct {S, \odot_1, \odot_2, \ldots, \odot_n}$
be automorphisms.
Then the composite of $\phi$ and $\psi$ is also an automorphism.
== Proof ==
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism.
By the definition of a composite mapping, $\psi \circ \phi$ is a mapping from $S$ into $S$.
Hence $\psi \circ \phi$ is an automorphism.
== Sources ==
* : $\S 1.2$
Category:Automorphisms (Abstract Algebra)
Composite of Bijection with Inverse is Identity Mapping

== Theorem ==
Let $f: S \to T$ be a bijection.
Then:
where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.
== Proof ==
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:
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:
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 $f \circ f^{-1} = I_T$.
== Also see ==
* Bijection iff Left and Right Inverse for the converse of this result.
== Sources ==
* : Introduction $\S 2$: Product sets, mappings
* : $\S 3.5$. Identity mappings: Example $52$
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.5$
* : Chapter $1$: Mappings: $\S 16$
* : $\S 1.14$: Composition of Functions: Exercise $6$
* : $\S 24.4$
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions
* 
* 
Category:Bijections
Category:Inverse Mappings
Category:Identity Mappings
Composite of Bijections is Bijection

== Theorem ==
Let $f$ and $g$ be mappings such that $\Dom f = \Cdm g$.
Then:
:If $f$ and $g$ are both bijections, then so is $f \circ g$
where $f \circ g$ is the composite mapping of $f$ with $g$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{J}$
* : Appendix: Elementary set and number theory
Category:Bijections
Category:Composite Mappings
Category:Composite of Bijections is Bijection
Composite of Bijections is Bijection/Proof 1

== Theorem ==
== Proof ==
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.
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations: $\text{(ii)}$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.10 \ (3)$
* : $\S 25.2$: Some further results and examples on mappings
* : $1$: Set Theory and Logic: $\S 2$: Functions
Category:Composite of Bijections is Bijection
Composite of Bijections is Bijection/Proof 2

== Theorem ==
== Proof ==
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.
== Sources ==
* : Chapter $2$: Maps and relations on sets: Corollary $2.16$
Category:Composite of Bijections is Bijection
Composite of Connected Relation is not necessarily Connected

== Theorem ==
Let $A$ be a set.
Let $\RR$ and $\SS$ be connected relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also connected.
== Proof ==
Recall the definition of composition of relations:
;Proof by Counterexample
:
Let $A = \set {a, b, c}$.
Let $\RR$ be defined as:
:$\RR = \set {\tuple {a, b}, \tuple {b, c}, \tuple {c, a} }$
Let $\SS$ be defined as $\RR^{-1}$, that is, the inverse of $\RR$:
:$\SS = \set {\tuple {b, a}, \tuple {c, b}, \tuple {a, c} }$
Both $\RR$ and $\SS$ can be seen to be connected.
Then we have:
:$\RR \circ \SS = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c} }$
and it is immediately apparent that $\RR \circ \SS$ is not connected.
Category:Composite Relations
Category:Connected Relations
Composite of Continuous Mappings at Point between Metric Spaces is Continuous at Point

== Theorem ==
Let $M_1 = \struct {X_1, d_1}, M_2 = \struct {X_2, d_2}, M_3 = \struct {X_3, d_3}$ be metric spaces.
Let $f: M_1 \to M_2$ be continuous at $a \in X_1$.
Let $g: M_2 \to M_3$ be continuous at $\map f a \in X_2$.
Then their composite $g \circ f: M_1 \to M_3$ is continuous at $a \in X_1$.
== Proof ==
Let $\epsilon \in \R_{>0}$.
The strategy is to find a $\delta \in \R_{>0}$ such that:
:$\map {d_1} {x, a} < \delta \implies \map {d_3} {\map g {\map f x}, \map g {\map f a} } < \epsilon$
As $g$ is continuous at $\map f a$:
:$\exists \eta \in \R_{>0}: \forall y \in X_2: \map {d_2} {y, \map f a} < \eta \implies \map {d_3} {\map g y, \map g {\map f a} } < \epsilon$
As $f$ is continuous at $a$:
:$\forall \eta \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in X_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \eta$
Hence:
:$\map {d_3} {\map g {\map f x}, \map g {\map f a} } < \epsilon$
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 3$: Continuity: Theorem $3.6$
Category:Metric Spaces
Category:Continuous Mappings on Metric Spaces
Category:Composite Mappings
Composite of Continuous Mappings between Metric Spaces is Continuous

== Theorem ==
Let $M_1 = \struct {X_1, d_1}$, $M_2 = \struct {X_2, d_2}$ and $M_3 = \struct {X_3, d_3}$ be metric spaces.
Let $f: M_1 \to M_2$ and $g: M_2 \to M_3$ be continuous mappings.
Then their composite $g \circ f: M_1 \to M_3$ is continuous.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Path-Connectedness
* : Chapter $2$: Metric Spaces: $\S 3$: Continuity: Corollary $3.7$
Category:Metric Spaces
Category:Continuous Mappings on Metric Spaces
Category:Composite Mappings
Category:Composite of Continuous Mappings between Metric Spaces is Continuous
Composite of Continuous Mappings between Metric Spaces is Continuous/Proof 2

== Theorem ==
== Proof ==
Let $f$ and $g$ be continuous mappings.
By definition:
:$f$ is continuous at $a \in X_1$ for all $a \in X_1$
:$g$ is continuous at $\map f a \in X_2$ for all $\map f a \in X_2$.
The result follows from Composite of Continuous Mappings at Point between Metric Spaces is Continuous at Point
Category:Composite of Continuous Mappings between Metric Spaces is Continuous
Composite of Continuous Mappings between Normed Vector Spaces is Continuous

== Theorem ==
Let $X, Y, Z$ be normed vector spaces.
Let $f : X \to Y$ and $g : Y \to Z$ be continuous mappings on $X$ and $Y$ respectively.
Let $g \circ f : X \to Z$ be a composite mapping where:
:$\forall x \in X : \map {\paren {g \circ f} } x := \map g {\map f x}$
Then $g \circ f$ is continuous on $X$.
== Proof ==
Let $W$ be open in $Z$.
$g$ is continuous on $Y$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {g^{-1} } W$ is open in $Y$.
$f$ is continuous on $X$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {f^{-1} } {\map {g^{-1} } W}$ is open in $X$.
By Inverse of Composite Relation:
:$\map {\paren {g \circ f}^{-1} } W = \map {f^{-1} } {\map {g^{-1} } W}$
Hence, $\map {\paren {g \circ f}^{-1} } W$ is continuous on $X$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $g \circ f$ is continuous on $X$.
== Also see ==
* Composite of Continuous Mappings is Continuous
* Composite of Continuous Mappings between Metric Spaces is Continuous
== Sources ==
* : Chapter $\S 2.2$: Continuous and linear maps. Continuous maps
Category:Continuous Mappings on Normed Vector Spaces
Category:Normed Vector Spaces
Category:Continuity
Category:Composite Mappings
Composite of Continuous Mappings is Continuous

== Theorem ==
Let $T_1, T_2, T_3$ be topological spaces.
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be continuous mappings.
Then the composite mapping $g \circ f: T_1 \to T_3$ is continuous.
=== Corollary ===
=== Continuity at a Point ===
== Proof ==
Let $U \in T_3$ be open in $T_3$.
As $g$ is continuous, $g^{-1} \sqbrk U \in T_2$ is open in $T_2$.
As $f$ is continuous, $f^{-1} \sqbrk {g^{-1} \sqbrk U} \in T_1$ is open in $T_1$.
By Preimage of Subset under Composite Mapping, $f^{-1} \sqbrk {g^{-1} \sqbrk U} = \paren {g \circ f}^{-1} \sqbrk U$.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Proposition $3.1.8$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Composite of Continuous Mappings is Continuous
Category:Continuous Mappings (Topology)
Category:Composite Mappings
Composite of Continuous Mappings is Continuous/Corollary

== Theorem ==
Let $T_1, T_2, T_3$ each be one of:
:metric spaces
:the complex plane
:the real number line
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be continuous mappings.
Then the composite mapping $g \circ f: T_1 \to T_3$ is continuous.
== Proof ==
These follow directly from:
* Real Number Line is Metric Space
* Complex Plane is Metric Space
* Metric Induces Topology
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.9$
* : $4.7$
Category:Composite of Continuous Mappings is Continuous
Category:Continuous Real Functions
Category:Continuous Complex Functions
Category:Continuous Mappings on Metric Spaces
Composite of Continuous Mappings is Continuous/Point

== Theorem ==
Let $T_1, T_2, T_3$ be topological spaces.
Let the mapping $f : T_1 \to T_2$ be continuous at $x$.
Let the mapping $g : T_2 \to T_3$ be continuous at $\map f x$.
Then the composite mapping $g \circ f : T_1 \to T_3$ is continuous at $x$.
== Proof ==
Let $N$ be any neighborhood of $\map {\paren {g \circ f} } x$.
By the definition of continuity at a point:
:there exists a neighborhood $L$ of $\map f x$ such that $g \sqbrk L \subseteq N$
and
:there exists a neighborhood $M$ of $x$ such that $f \sqbrk M \subseteq L$.
Thus $\paren {g \circ f} \sqbrk M \subseteq g \sqbrk L \subseteq N$, as desired.
== Sources ==
* 
* 
Category:Composite of Continuous Mappings is Continuous
Category:Composite Mappings
Category:Continuous Mappings (Topology)
Composite of Endomorphisms is Endomorphism

== Theorem ==
Let $\struct {S, \odot_1, \odot_2, \ldots, \odot_n}$ be an algebraic structure.
Let:
:$\phi: \struct {S, \odot_1, \odot_2, \ldots, \odot_n} \to \struct {S, \odot_1, \odot_2, \ldots, \odot_n}$
:$\psi: \struct {S, \odot_1, \odot_2, \ldots, \odot_n} \to \struct {S, \odot_1, \odot_2, \ldots, \odot_n}$
be endomorphisms.
Then the composite of $\phi$ and $\psi$ is also an endomorphism.
== Proof ==
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism.
From the definition of composition of mappings, $\psi \circ \phi$ is a mapping from $S$ into $S$.
Hence $\psi \circ \phi$ is an endomorphism.
== Sources ==
* : $\S 1.2$
Category:Endomorphisms
Composite of Epimorphisms is Epimorphism

== Theorem ==
Let:
:$\struct {S_1, \circ_1, \circ_2, \ldots, \circ_n}$
:$\struct {S_2, *_1, *_2, \ldots, *_n}$
:$\struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$
be algebraic structures.
Let:
:$\phi: \struct {S_1, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {S_2, *_1, *_2, \ldots, *_n}$
:$\psi: \struct {S_2, *_1, *_2, \ldots, *_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$
be epimorphisms.
Then the composite of $\phi$ and $\psi$ is also an epimorphism.
== Proof ==
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism.
From Composite of Surjections is Surjection, $\psi \circ \phi$ is a surjection.
An epimorphism is a surjective homomorphism.
Hence $\psi \circ \phi$ is an epimorphism.
== Sources ==
* : $\S 1.2$
Category:Epimorphisms (Abstract Algebra)
Category:Composite Mappings
Composite of Evaluation Mapping and Projection

== Theorem ==
Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Let:
:$pr_i : Y \to Y_i$ denote the $i$th projection on $Y$
Then:
:$\forall i \in I: pr_i \circ f = f_i$
== Proof ==
By definition of projection:
From Equality of Mappings:
:$\forall i \in I : pr_i \circ f = f_i$
Category:Evaluation Mappings (Topological Spaces)
Composite of Group Epimorphisms is Epimorphism

== Theorem ==
Let:
:$\struct {G_1, \odot}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be groups.
Let:
:$\phi: \struct {G_1, \odot} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be (group) epimorphisms.
Then the composite of $\phi$ and $\psi$ is also a (group) epimorphism.
== Proof ==
A group epimorphism is a group homomorphism which is also a surection.
From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a group homomorphism.
From Composite of Surjections is Surjection, $\psi \circ \phi$ is a surection.
== Sources ==
* : Chapter $\text{II}$: Groups: Factoring Morphisms
Category:Group Epimorphisms
Composite of Group Homomorphisms is Homomorphism

== Theorem ==
Let:
:$\struct {G_1, \circ}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be groups.
Let:
:$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be homomorphisms.
Then the composite of $\phi$ and $\psi$ is also a homomorphism.
== Proof 1 ==
== Proof 2 ==
Category:Group Homomorphisms
Category:Composite of Group Homomorphisms is Homomorphism
Composite of Group Homomorphisms is Homomorphism/Proof 2

== Theorem ==
== Proof ==
So as to alleviate possible confusion over notation, let the composite of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$.
Then what we are trying to prove is denoted:
:$\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ is a homomorphism.
To prove the above is the case, we need to demonstrate that the morphism property is held by $\circ$ under $\psi \bullet \phi$.
We take two elements $x, y \in G_1$, and put them through the following wringer:
Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by $\circ$ under $\psi \bullet \phi$.
Thus $\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ is indeed a homomorphism.
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Lemma $\text{(iv)}$
* : Chapter $\text{II}$: Groups: Factoring Morphisms
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 60$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $3 \ \text{(i)}$
Category:Composite of Group Homomorphisms is Homomorphism
Composite of Group Isomorphisms is Isomorphism

== Theorem ==
Let $\struct {G_1, \circ}$, $\struct {G_2, *}$ and $\struct {G_3, \oplus}$ be groups.
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ and $\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$ be group isomorphisms.
Then the composite of $\psi$ with $\phi$ is also a group isomorphism.
== Proof ==
A group isomorphism is a group homomorphism which is also a bijection.
From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a group homomorphism.
From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection.
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Lemma $\text{(iv)}$
* : Chapter $\text{II}$: Groups: Factoring Morphisms
Category:Group Isomorphisms
Composite of Group Monomorphisms is Monomorphism

== Theorem ==
Let:
:$\struct {G_1, \circ}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be groups.
Let:
:$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be monomorphisms.
Then the composite of $\phi$ and $\psi$ is also a monomorphism.
== Proof ==
A monomorphism is a homomorphism which is also an injection.
From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a homomorphism.
From Composite of Injections is Injection, $\psi \circ \phi$ is an injection.
== Sources ==
* : Chapter $\text{II}$: Groups: Factoring Morphisms
Category:Group Monomorphisms
Composite of Homeomorphisms between Metric Spaces is Homeomorphism

== Theorem ==
Let $M_1, M_2, M_3$ be metric spaces. 
Let $f: M_1 \to M_2$ and $g: M_2 \to M_3$ be homeomorphisms.
Then $g \circ f: M_1 \to M_3$ is also a homeomorphism.
== Proof ==
By definition of homeomorphism, $f$ and $g$ are both bijections.
From Composite of Bijections is Bijection it follows that $g \circ f$ is also a bijection.
Similarly, also by definition of homeomorphism, $f$ and $g$ are both continuous mappings.
From Composite of Continuous Mappings on Metric Spaces is Continuous it follows that $g \circ f$ is also a continuous mapping.
From Inverse of Bijection is Bijection, $f^{-1}: M_2 \to M_1$ and $g^{-1}: M_3 \to M_2$ are also both bijections.
From Composite of Bijections is Bijection it follows that $f^{-1} \circ g^{-1} : M_3 \to M_1$ is also a bijection.
By definition of homeomorphism, $f^{-1}$ and $g^{-1}$ are both continuous mappings.
Again from Composite of Continuous Mappings on Metric Spaces is Continuous it follows that $f^{-1} \circ g^{-1}$ is also a continuous mapping.
Hence the result, from definition of homeomorphism.
Category:Homeomorphisms (Metric Spaces)
Composite of Homeomorphisms is Homeomorphism

== Theorem ==
Let $T_1, T_2, T_3$ be topological spaces. 
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be homeomorphisms.
Then $g \circ f: T_1 \to T_3$ is also a homeomorphism.
== Proof ==
By definition of homeomorphism, $f$ and $g$ are both bijections.
From Composite of Bijections is Bijection it follows that $g \circ f$ is also a bijection.
By definition of homeomorphism, $f$ and $g$ are both continuous mappings.
From Composite of Continuous Mappings is Continuous it follows that $g \circ f$ is also a continuous mapping.
From Inverse of Bijection is Bijection, $f^{-1}: T_2 \to T_1$ and $g^{-1}: T_3 \to T_2$ are also both bijections.
From Composite of Bijections is Bijection it follows that $f^{-1} \circ g^{-1}$ is also a bijection.
By definition of homeomorphism, $f^{-1}$ and $g^{-1}$ are both continuous mappings.
Again from Composite of Continuous Mappings is Continuous it follows that $f^{-1} \circ g^{-1}$ is also a continuous mapping.
Hence the result, from definition of homeomorphism.
Category:Homeomorphisms (Topological Spaces)
Composite of Homomorphisms is Homomorphism/Algebraic Structure

== Theorem ==
Let:
:$\struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n}$
:$\struct {S_2, *_1, *_2, \ldots, *_n}$
:$\struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$
be algebraic structures.
Let:
:$\phi: \struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n} \to \struct {S_2, *_1, *_2, \ldots, *_n}$
:$\psi: \struct {S_2, *_1, *_2, \ldots, *_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$
be homomorphisms.
Then the composite of $\phi$ and $\psi$ is also a homomorphism.
== Proof ==
Let $\psi \circ \phi$ denote the composite of $\phi$ and $\psi$.
Then what we are trying to prove is denoted:
:$\paren {\psi \circ \phi}: \struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$ is a homomorphism.
To prove the above is the case, we need to demonstrate that the morphism property is held by each of the operations $\otimes_1, \otimes_2, \ldots, \otimes_n$ under $\psi \circ \phi$.
Let $\otimes_k$ be one of these operations.
We take two elements $x, y \in S_1$, and put them through the following wringer:
Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by $\otimes_k$ under $\psi \circ \phi$.
As this holds for any arbitrary operation $\otimes_k$ in $\struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n}$, it follows that it holds for all of them.
Thus $\paren {\psi \circ \phi}: \struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$ is indeed a homomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.4$
* : $\S 1.2$
Category:Composite of Homomorphisms is Homomorphism
Category:Homomorphisms (Abstract Algebra)
Category:Composite Mappings
Composite of Homomorphisms is Homomorphism/R-Algebraic Structure

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let:
:$\struct {S_1, \odot_1, \ldots, \odot_n}$
:$\struct {S_2, {\odot'}_1, \ldots, {\odot'}_n}$
:$\struct {S_3, {\odot' '}_1, \ldots, {\odot' '}_n}$
be algebraic structures each with $n$ operations.
Let:
:$\struct {S_1, *_1}_R$
:$\struct {S_2, *_2}_R$
:$\struct {S_3, *_3}_R$
be $R$-algebraic structures.
Let:
:$\phi: \struct {S_1, *_1}_R \to \struct {S_2, *_2}_R$
:$\psi: \struct {S_2, *_2}_R \to \struct {S_3, *_3}_R$
be homomorphisms.
Then the composite of $\phi$ and $\psi$ is also a homomorphism.
== Proof ==
What we are trying to prove is that $\paren {\psi \circ \phi}: \struct {S_1, *_1}_R \to \struct {S_3, *_3}_R$ is a homomorphism.
That is:
:$(1): \quad \forall k \in \closedint 1 n: \forall x, y \in S_1: \map {\paren {\psi \circ \phi} } {x \odot_k y} = \map {\paren {\psi \circ \phi} } x {\odot' '}_k \map {\paren {\psi \circ \phi} } y$
:$(2): \quad \forall x \in S: \forall \lambda \in R: \map {\paren {\psi \circ \phi} } {\lambda *_1 x} = \lambda *_3 \map \phi x$
where $\closedint 1 n = \set {1, 2, \ldots, n}$ denotes an integer interval.
To prove the above is the case, we need to demonstrate that the morphism property is held by each of the operations $\odot_1, \odot_2, \ldots, \odot_n$ under $\psi \circ \phi$.
Let $\odot_k$ be one of these operations.
We take two elements $x, y \in S_1$, and put them through the following wringer:
Hence the morphism property is held by $\odot_k$ under $\psi \circ \phi$.
As $\odot_k$ is an arbitrary operation in $\struct {S_1, \odot_1, \ldots, \odot_n}$, it follows that morphism property is held by all of $\odot_1, \ldots, \odot_n$.
Thus $\paren {\psi \circ \phi}: \struct {S_1, *_1}_R \to \struct {S_3, *_3}_R$ satisfies $(1)$ above.
We take an element $x \in S_1$, and apply the following:
Hence the morphism property is held by $\odot_k$ under $\psi \circ \phi$.
As $\odot_k$ is an arbitrary operation in $\struct {S_1, \odot_1, \ldots, \odot_n}$, it follows that morphism property is held by all of $\odot_1, \ldots, \odot_n$.
Thus $\paren {\psi \circ \phi}: \struct {S_1, *_1}_R \to \struct {S_3, *_3}_R$ satisfies $(2)$ above.
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.1$
Category:Composite of Homomorphisms is Homomorphism
Category:Homomorphisms (Abstract Algebra)