How Is u_alpha an Injective Mapping in Proposition 2.1.4?

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need help with some aspects of the proof of Proposition 2.1.4 ...

Proposition 2.1.4 and its proof read as follows:View attachment 8039
In the above proof by Paul Bland we read the following:

" ... ... Since \(\displaystyle p_\alpha u_\alpha = \text{ id}_{ M_\alpha }\), we have that \(\displaystyle u_\alpha\) is an injective mapping and that \(\displaystyle p_\alpha\) is surjective ... ... "Can someone please explain exactly how/why \(\displaystyle u_\alpha\) is an injective mapping ... ?Help will be appreciated ...

Peter

 

[steenis]

How do you prove that an $R$-map in injective? Make a start with the proof that $u_\alpha$ is injective.

 

[Math Amateur]

How do you prove that an $R$-map in injective? Make a start with the proof that $u_\alpha$ is injective.

Hi Steenis ...

To show that \(\displaystyle u_\alpha\) is injective ...

... we assume that \(\displaystyle u_\alpha (a) = u_\alpha (b)\) ...

... then we need to show that \(\displaystyle a = b\) ...So ... let \(\displaystyle u_\alpha (a) = u_\alpha (b)\) ... Now \(\displaystyle p_\alpha u_\alpha = \text{id}_{ M_\alpha }\) ...

... so ... \(\displaystyle p_\alpha u_\alpha (a) = \text{id}_{ M_\alpha } (a) = a\) ... ... ... ... (1)

... and ... \(\displaystyle p_\alpha u_\alpha (b) = \text{id}_{ M_\alpha } (b) = b\) ... ... ... ... (2)... and we also have \(\displaystyle p_\alpha^{ -1} p_\alpha u_\alpha (a) = p_\alpha^{ -1} \text{id}_{ M_\alpha } (a) = p_\alpha^{ -1} ( a )
\)

\(\displaystyle \Longrightarrow u_\alpha (a) = p_\alpha^{ -1} ( a )
\)... and similarly we have ... \(\displaystyle u_\alpha (b) = p_\alpha^{ -1} ( b )\)
... ... do not seem to be progressing toward objective of demonstrating that \(\displaystyle a = b\) ...
Can you help?

Peter***EDIT***oh! ... (1) and (2) and our assumption give us the answer ...Putting \(\displaystyle u_\alpha (b) = u_\alpha (a)\) into (2) gives

\(\displaystyle p_\alpha u_\alpha (b) = p_\alpha u_\alpha (a) = a = \text{id}_{ M_\alpha } (b) = b\)

and QED!

Is that correct ...?

Peter

 

[steenis]

YES, that is correct.

There is a more general rule:
Let $A$, $B$, and $C$ be sets, and $f:A \longrightarrow B$, $g:B \longrightarrow C$, and $h=g \circ f:A \longrightarrow C$ (set-) functions, then if $h$ is bijective then $g$ is surjective and $f$ is injective.

This also applies to $R$-maps:
$A$, $B$, and $C$ are $R$-modules, $f:A \longrightarrow B$, $g:B \longrightarrow C$, and $h=g \circ f:A \longrightarrow C$ are $R$-maps, then if $h$ is an isomorphism, then $g$ is an epimorphism and $f$ is an monomorphism.

The proofs are little different than your proof, but easy, you can try it.

 

[Math Amateur]

YES, that is correct.

There is a more general rule:
Let $A$, $B$, and $C$ be sets, and $f:A \longrightarrow B$, $g:B \longrightarrow C$, and $h=g \circ f:A \longrightarrow C$ (set-) functions, then if $h$ is bijective then $g$ is surjective and $f$ is injective.

This also applies to $R$-maps:
$A$, $B$, and $C$ are $R$-modules, $f:A \longrightarrow B$, $g:B \longrightarrow C$, and $h=g \circ f:A \longrightarrow C$ are $R$-maps, then if $h$ is an isomorphism, then $g$ is an epimorphism and $f$ is an monomorphism.

The proofs are little different than your proof, but easy, you can try it.

Thanks Steenis ...

Peter