1019: For Linear Map Between Modules with Bases, if Its Restriction on Domain Basis Is Bijection onto Codomain Basis, Map Is 'Modules - Linear Morphisms' Isomorphism

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description/proof of that for linear map between modules with bases, if its restriction on domain basis is bijection onto codomain basis, map is 'modules - linear morphisms' isomorphism

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Topics

About: module

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of basis of module.
• The reader knows a definition of linear map.
• The reader knows a definition of bijection.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
• The reader admits the proposition that any bijective linear map between any modules is a 'modules - linear morphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that for any linear map between any modules with bases, if its restriction on any domain basis is a bijection onto any codomain basis, the map is a 'modules - linear morphisms' isomorphism.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the rings }\}$
$M_1$: $\in \{\text{ the }R\text{ modules }\}$
$M_2$: $\in \{\text{ the }R\text{ modules }\}$
$J_1$: $\in \{\text{ the possibly uncountable index sets }\}$
$J_2$: $\in \{\text{ the possibly uncountable index sets }\}$
$B_1$: $=\{b_{1,j}|j\in J_1\}$, $\in \{\text{ the bases for }{M}_1\}$
$B_2$: $=\{b_{2,j}|j\in J_2\}$, $\in \{\text{ the bases for }{M}_2\}$
$f$: $:M_1\to M_2$
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Statements:
$f(B_1)=B_2\wedge f{|}_{B_1}:B_1\to B_2\in \{\text{ the bijections }\}$
$⟹$
$f\in \{\text{ the 'modules - linear morphisms' isomorphisms }\}$
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2: Proof

Whole Strategy: Step 1: see that $f$ is injective; Step 2: see that $f$ is surjective; Step 3: conclude the proposition.

Step 1:

Let us see that $f$ is injective.

Let $m,m^{\prime }\in M_1$ be any such that $m\ne m^{\prime }$.

There are a finite $S\subseteq J_1$ such that $m=\sum _{j\in S}{m}^j{b}_{1,j}$ and a finite $S^{\prime }\subseteq J_1$ such that $m^{\prime }=\sum _{j\in S^{\prime }}{m}^{\prime j}{b}_{1,j}$.

Let us take $S\cup S^{\prime }$ and for each $j\in S\cup S^{\prime }$, ${\tilde{m}}^j=m^j$ for $j\in S$ and ${\tilde{m}}^j=0$ for $j\notin S$ and ${\widetilde{m^{\prime }}}^j=m^{\prime j}$ for $j\in S^{\prime }$ and ${\widetilde{m^{\prime }}}^j=0$ for $j\notin S^{\prime }$.

$m=\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_{1,j}$ and $m^{\prime }=\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_{1,j}$.

As $m\ne m^{\prime }$, ${\tilde{m}}^l\ne {\widetilde{m^{\prime }}}^l$ for a $l\in S\cup S^{\prime }$.

$f(m)=f(\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_{1,j})=\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^{j}f(b_{1,j})$, because $f$ is linear.

$f(m^{\prime })=f(\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_{1,j})=\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^{j}f(b_{1,j})$, because $f$ is linear.

As $f(B_1)=B_2$, each $f(b_{1,j})$ is an element of $B_2$, and as $f{|}_{B_1}$ is bijective, $\{f(b_{1,j})|j\in S\cup S^{\prime }\}$ is distinct.

As ${\tilde{m}}^l\ne {\widetilde{m^{\prime }}}^l$, $f(m)=\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^{j}f(b_{1,j})\ne \sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^{j}f(b_{1,j})=f(m^{\prime })$, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

Step 2:

Let $m\in M_2$ be any.

There is a finite $S\subseteq J_2$ such that $m=\sum _{j\in S}{m}^j{b}_{2,j}$.

As $f{|}_{B_1}:B_1\to B_2$ is bijective, there is the inverse, ${f{|}_{B_1}}^{-1}:B_2\to B_1$. So, ${f{|}_{B_1}}^{-1}(b_{2,j})\in B_1$.

Let us think of $\sum _{j\in S}{m}^j{f{|}_{B_1}}^{-1}(b_{2,j})\in M_1$.

$f(\sum _{j\in S}{m}^j{f{|}_{B_1}}^{-1}(b_{2,j}))=\sum _{j\in S}{m}^{j}f({f{|}_{B_1}}^{-1}(b_{2,j}))$, because $f$ is linear, $=\sum _{j\in S}{m}^j{b}_{2,j}=m$.

So, $f$ is surjective.

Step 3:

By the proposition that any bijective linear map between any modules is a 'modules - linear morphisms' isomorphism, $f$ is a 'modules - linear morphisms' isomorphism.

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References

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