Undergrad Split Epimorphisms .... Bland Proposition 3.2.4 ....

[Math Amateur]

I am reading Paul E. Bland's book "Rings and Their Modules" ...

Currently I am focused on Section 3.2 Exact Sequences in ModR" role="presentation">ModR ... ...

I need some help in order to fully understand Proposition 3.2.4 ...

Proposition 3.2.4 reads as follows:

Can someone please explain exactly how Proposition 3.2.3 establishes Proposition 3.2.4 ...
Help will be much appreciated ...

Peter

 

[andrewkirk]

Start with the premises of 3.2.4 about $g,g'$.

Then, from the author's comment after Definition 3.2.2 we see that $g'$ is a split monomorphism with monomorphism-splitting map $g$.

Hence by Proposition 3.2.3, the range $M$ of the split monomorphism $g'$ can be written as a direct sum of the split monomorphism's image ($\textrm{Im}\ g'$) and the Kernel of the monomorphism-splitting map $g$.

I find it helps to attach the word 'epimorphism-' or 'monomorphism-' as a prefix to the words 'splitting map' in order to avoid confusion as to whether the map in question is splitting an epimorphism or a monomorphism.

Are you able to satisfy yourself as to the truth of the author's comment after Definition 3.2.2?

 

[Math Amateur]

Hi Andrew ... thanks for the help ...

Regarding the authors comments after Definition 3.2.2 ...

Firstly, if $f$ is a split monomorphism with splitting map $f'$... then any point/element of $M_1$, say $a$, must be mapped to itself: ... hence $f'$ must be surjective ... that is $f'$ is an epimorphism ...

Secondly if $g$ is a split epimorphism with splitting map $g'$... then any point $a \in M_2$ will be mapped to $g'(a)$ ... and no other point of $M_2$, say $b$, may be mapped to $g'(a)$ ... otherwise ... $g$ must map the point $g'(a)$ back to both $a$ and $b$ ... which is impossible since $g$ is a map ... therefore $g'$ is injective ... that is, $g'$ is a monomorphism ... ...

Peter

 

[andrewkirk]

Yes. A neat way to put the second one is to let $a,b\in M_2$ and assume $g'(a)=g'(b)$. Then we have

$$a=id_{M_2}(a) = (g\circ g')(a) = g(g'(a)) = g(g'(b)) = (g\circ g')(b) = id_{M_2}(b) = b$$

which gives us the injectivity property of $g'$.

 

[steenis]

The key to your question is this lemma:

If $f:M \to N$ and $g:N \to M$ are R-maps such that $g \circ f = 1_M$ then $f$ is an R-monomorphism and $g$ is an R-epimorphism. You may prove that.
Here we have a sequence

$M\overset{f}{\rightarrow}N\overset{g}{\rightarrow}M$

of R-maps such that $g \circ f = 1_M$.In proposition 3.2.3. we have this sequence

$M_1\overset{f}{\rightarrow}M\overset{f’}{\rightarrow}M_1$

of R-maps such that $f’ \circ f = 1_{M_1}$.

By the lemma, $f$ is a monomorphism and we have $f’ \circ f = 1_{M_1}$: definition 3.2.2. calls $f$ the split monomorphism and calls $f’$ the splitting map for $f$.

Proposition 3.2.3 says $M = \text{im } f \oplus \text{ker } f’$


Can you see this:

By the lemma, $f’$ is an epimorphism and we have $f’ \circ f = 1_{M_1}$: but now definition 3.2.2. calls $f’$ the split epimorphism and $f$ the splitting map for $f’$.

Proposition 3.2.3 says $M = \text{im } f \oplus \text{ker } f’$This is exactly the situation we have in proposition 3.2.4. here we have this sequence

$M_2\overset{g’}{\rightarrow}M\overset{g}{\rightarrow}M_2$

of R-maps such that $g \circ g’ = 1_{M_2}$: $g$ the split epimorphism and $g’$ the splitting map for $g$.

So change $f$ into $g’$, and $f’$ into $g$, and $M_1$ into $M_2$ and proposition 3.2.3 becomes proposition 3.2.4. And the conclusion is $M = \text{im } g’ \oplus \text{ker } g$