A little problem on exact sequences.

[steenis]

I have this little problem on exact sequences, I want to check with you.

We have two R-maps $f:A\to B$ and $g:B\to C$ of left R-modules.
And we have an isomorphism $B/\mbox{im} f\cong C$ "induced by g" .
Then prove that the sequence $S:A\to _f B\to _g C\to 0$ is right-exact, i.e.,
$\mbox{im} f = \ker g$ and $g$ is surjective.

"Induced by g", I do not know what this exactly means in this context.
I think it means that the isomorphism is given by
$B/\mbox{im} f \to C:b+\mbox{im} f \mapsto g(b)$.

Your thoughts, please.

 

[Deveno]

I have this little problem on exact sequences, I want to check with you.

We have two R-maps $f:A\to B$ and $g:B\to C$ of left R-modules.
And we have an isomorphism $B/\mbox{im} f\cong C$ "induced by g" .
Then prove that the sequence $S:A\to _f B\to _g C\to 0$ is right-exact, i.e.,
$\mbox{im} f = \ker g$ and $g$ is surjective.

"Induced by g", I do not know what this exactly means in this context.
I think it means that the isomorphism is given by
$B/\mbox{im} f \to C:b+\mbox{im} f \mapsto g(b)$.

Your thoughts, please.

I think so, too.

If such an $R$-module $R$-map exists, of necessity it must map $\text{im f} \to 0_C$ ($R$-maps send $0 \to 0$).

But then this means $\overline{g}:B/(\text{im f}) \to C$ given by $\overline{g}(b + \text{im }f) = g(b)$ is well-defined, since it is constant on cosets.

If such a $\overline{g}$ is an isomorphism, then surely $g$ is surjective, since we have, for any $c \in C$, some coset:

$b + \text{im }f \in B/(\text{im }f)$ with $\overline{g}(b + \text{im }f) = c$, that is: $g(b) = c$.

Clearly, $\text{im }f \subseteq \text{ker }\overline{g}$, and since $\overline{g}$ is injective, we have:

$g(b) = 0_C \implies b \in \text{im }f$, proving that $\text{ker }g = \text{im }f$.

 

[steenis]

Thank you, Deveno.

For clarity:
$\mbox{im }f \subset \ker (g)$ because if $y\in \mbox{im } f$ then $\overline y =0$ and $g(y)= \overline {g} (\overline {y})=0$

In the meanwhile, I posted this question on MSE: abstract algebra - A statement on exact sequences of R-modules. - Mathematics Stack Exchange
That produced two interesting ansers.

I posed this question because I want to understand the answer of this MSE-post: commutative algebra - Exact sequence of $A$-modules - Mathematics Stack Exchange

 

[steenis]

@Deveno. Rethinking your answer, I noticed that I do not understand this:

But then this means $\overline{g}:B/(\text{im f}) \to C$ given by $\overline{g}(b + \text{im }f) = g(b)$ is well-defined, since it is constant on cosets.

Maybe you can clarify this a little more.I think you do not know enough to prove the well-definition of $\overline g$ and you must assume it, because it is given that it is an R-map.

 

[Deveno]

@Deveno. Rethinking your answer, I noticed that I do not understand this:
Maybe you can clarify this a little more.I think you do not know enough to prove the well-definition of $\overline g$ and you must assume it, because it is given that it is an R-map.

The answer posted by Zev Conoles on mathstackexchange is a good one, and relevant, here. When we say $\overline{g}$ is induced by $g$, we mean two things:

1. $\text{im f} \subseteq \text{ker }g$ -this is what ensures $\overline{g}$ is well-defined.

2. $\overline{g}(b + \text{im }f) = g(b)$.

In other words, if $\pi$ is the canonical map $B \to B/(\text{im f})$, we have $g = \overline{g}\circ\pi$. Often, one uses the expression "$g$ factors through $\pi$", and thus the induced map is "the other factor".

You are partially correct in one respect: the fact that $\overline{g}$ is well-defined is *equivalent* to #1. above: if $g$ is constant on cosets of a submodule $M$ of $B$, it induces a well-defined mapping: $B/M \to C$, and if it does induce such a mapping, it must be constant on cosets of $M$.

 

[steenis]

I must say that I am not very bright, I did mention that $\mbox{im } f \subset \ker g$, and still did not see that that is the key to the well-definition of $\overline g$, ...

Thank you, Deveno, your post is a relevant addition.
You may go to part 2, :).

 

[Deveno]

I must say that I am not very bright, I did mention that $\mbox{im } f \subset \ker g$, and still did not see that that is the key to the well-definition of $\overline g$, ...

Thank you, Deveno, your post is a relevant addition.
You may go to part 2, :).

Don't sell yourself short, your posts evidence insightful thinking into the ideas, and an clear ability to focus on "weak links" in chains of reasoning. I find your contributions to the site welcome and refreshing.

 

[steenis]

Thank you, Deveno, you are very kind !