MHB Exact Sequences - Split Sequences and Splitting Homomorphisms

[Math Amateur]

I am reading Dummit and Foote Section 10.5 Exact Sequences - Projective, Injective and Flat Modules.

I need some help in understanding D&F's proof of Proposition 25, Section 10.5 (page 384) concerning split sequences.

Proposition 25 and its proof are as follows:

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Proposition 25. The short exact sequence $$ 0 \longrightarrow A \stackrel{\psi}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C \longrightarrow 0 $$ of R-modules is split if and only if there is an R-module homomorphism $$ \mu \ : \ C \to B $$ such that $$ \phi \circ \mu $$ is the identity map id on C. Similarly, the short exact sequence $$ 1 \longrightarrow A \stackrel{\psi}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C \longrightarrow 1 $$ of groups is split if and only if there is a group homomorphism $$ \mu \ : \ C \to B $$ such that $$ \phi \circ \mu $$ is the identity map id on C.

Proof: This follows directly from the definitions: if $$ \mu $$ is given then define $$ c' = \mu (C) \subseteq B $$ and if C' is given then define $$ \mu = \phi^{-1} \ : \ C \cong C' \subseteq B $$.

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Now I need some help with the proof ... to be specific ... suppose we are given $$ \mu \ : \ C \to B $$ such that $$ \phi \circ \mu = id $$ on C ... ... how, then, do we show that the short exact sequence is split ... that is, how do we show that:

$$ B = \psi (A) \oplus C' $$ ... ... ... ... (1)

for the submodule C' of B?

Following D&F we put $$ C' = \mu (C) \subseteq B $$

... ... but how then do we establish (1)?

I would appreciate some help.

Peter

 

[ThePerfectHacker]

We say that the short-exact sequence,
$$ 0 \to A \to_f B \to_g C \to 0 $$
Is split iff there is a $h:C\to B$ such that $hg = 1$.

We want to construct two submodule of $B$, call them $B_1,B_2$ such that:
(i) $B = B_1\oplus B_2$ i.e. every element of $B$ is uniquely written as $b_1+b_2$ where $b_1\in B_1,b_2\in B_2$
(ii) $B_1 = f(A)$ and $g(B_2) = C$ with $g$ being an isomorphism between $B_2$ and $C$ ($f$ is clearly an isomorphism).

To show that $B$ can be decomposed in such a way when we have a split short-exact sequence we let $B_1 = f(A)$ and $B_2 = h(C)$. We claim that $B_1\oplus B_2 = B$.

To show that we need to show that every element decomposes as a unique sum with $B_1,B_2$. Let $b\in B$, define $b_2 = h(g(b))\in B_2$. Now set $b_1 = b - b_2$. Note that we have $b_1 + b_2 = b$. We know that $b_2 \in B_2$ but why is $b_1\in B_1$? To see why observe that $g(b_1) = g(b-b_2) = g(b) - g(h(g(b))) = g(b) - g(b) = 0$. Thus, $b_1 \in \ker g$ but $\ker g = \text{im} f$ (definition of exact) which means that $b_1 \in f(A) = B_1$.

I guess it remains to show why this decomposition is unique to be finished.

 

[Math Amateur]

We say that the short-exact sequence,
$$ 0 \to A \to_f B \to_g C \to 0 $$
Is split iff there is a $h:C\to B$ such that $hg = 1$.

We want to construct two submodule of $B$, call them $B_1,B_2$ such that:
(i) $B = B_1\oplus B_2$ i.e. every element of $B$ is uniquely written as $b_1+b_2$ where $b_1\in B_1,b_2\in B_2$
(ii) $B_1 = f(A)$ and $g(B_2) = C$ with $g$ being an isomorphism between $B_2$ and $C$ ($f$ is clearly an isomorphism).

To show that $B$ can be decomposed in such a way when we have a split short-exact sequence we let $B_1 = f(A)$ and $B_2 = h(C)$. We claim that $B_1\oplus B_2 = B$.

To show that we need to show that every element decomposes as a unique sum with $B_1,B_2$. Let $b\in B$, define $b_2 = h(g(b))\in B_2$. Now set $b_1 = b - b_2$. Note that we have $b_1 + b_2 = b$. We know that $b_2 \in B_2$ but why is $b_1\in B_1$? To see why observe that $g(b_1) = g(b-b_2) = g(b) - g(h(g(b))) = g(b) - g(b) = 0$. Thus, $b_1 \in \ker g$ but $\ker g = \text{im} f$ (definition of exact) which means that $b_1 \in f(A) = B_1$.

I guess it remains to show why this decomposition is unique to be finished.

Thanks ThePerfectHacker, really appreciate the help ... ...

Just a couple of clarifications:

You write:

"We say that the short-exact sequence,
$$ 0 \to A \to_f B \to_g C \to 0 $$
Is split iff there is a $h:C\to B$ such that $hg = 1$. "Isn't the condition in this proposition $$ gh = 1 $$? - see Dummit and Foote, page 384 (see attachment)Another issue I am concerned about is that we say in (ii) above that we require $B_1 = f(A)$ and $g(B_2) = C$ with $g$ being an isomorphism between $B_2$ and $C$ ... ... ... but ... we do not seem anywhere to have shown that the homomorphism $g$ is an isomorphism between $B_2$ and $C$ ... ...A third issue ... where I may be misunderstanding what is going on ... is as follows ... in your post you write:

" ... ... define $$ b_2 = h(g(b)) $$ ... ... "

... but since hg = 1 we have $$ h(g(b)) = b $$ ... but then this does not support us writing "set $$ b_1 = b - b_2 $$ ... can you clarify?Can you help with my concerns above?

Peter

 

[Deveno]

There is a "hidden" observation underlying this:

If $g \circ h$ is bijective, then $g$ is injective, and $h$ is surjective. Before you go any further, convince yourself this is true.

Given the short exact sequence:

$0 \to A \to B \to C \to 0$

it is natural to regard $A$ as an $R$-submodule of $B$, and $C$ as $B/A$ (this is only true "up to isomorphism", but so?).

If this sequence is split, then we have $B/A$ isomorphic to an $R$-submodule of $A$, since the identity map on $C$ is certainly bijective.

So the claim we are proving is essentially equivalent to:

$B \cong A \oplus B/A$

that is, any element $b \in B$ is uniquely determined by first finding which coset $b + A$ it lies in, and then adding a specific element of $A$ to it. Stated this way, it seems almost obvious (recall that the cosets partition $B$).

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Yes, TPH does have it backwards, the condition is that $gh = 1$, but this does not change his proof that $g(b_1) = 0$:

$g(b_1) = g(b - b_2) = g(b) - g(b_2) = g(b) - g(h(g(b))) = g(b) - (g \circ h)(g(b)) = g(b) - g(b) = 0$

My remarks above indicates why $h$ is an isomorphism between $C$ and $h(C)$ (limiting its co-domain to its image forces surjectivity, and I say above why it is injective). It may not be clear why $g$ is likewise an isomorphism, when restricted to $h(C)$. Clearly, $g$ is surjective on this set, since we have:

$c = (gh)(c) = g(h(c))$ so any $c \in C$ has the pre-image (under $g$) of $h(c)$.

Now suppose we have $b_1,b_2 \in h(C)$, with $g(b_1) = g(b_2)$. We can write:

$b_1 = h(c_1), b_2 = h(c_2)$ since these are both in the image of $h$, so:

$g(b_1) = g(b_2) \implies g(h(c_1)) = g(h(c_2)) \implies (gh)(c_1) = (gh)(c_2) \implies c_1 = c_2$.

This, in turn, means that $b_1 = h(c_1) = h(c_2) = b_2$, so $g$ is injective.

You also ask if we don't have:

$(hg)(b) = b$

In general, NO: $g$ is ONLY injective when restricted to $h(C)$ (it is "super-duper non-injective" on the image of $A$ under $f$), and there is no reason to suppose that $b$ lies within the image of $h$ (part of it does, the "$C$" part, the other part, the "$A$" part, gets shrunk by $g$ to nuffin').

***********

Let's look at this in another light:

We have 3 "main maps" in a split short exact sequence:

$f:A \to B$, injective.
$g: B \to C$, surjective
$h: C \to B$, injective.

Note that this gives us a PAIR of injective $R$-module homomorphisms into $B$.

What is $\text{im }h \cap \text{im }f$?

Since the sequence is exact, $\text{im }f = \text{ker g}$.

If $b \in \text{im }h \cap \text{ker }g$, then:

$g(b) = 0$ and $h(c) = b$ whence:

$c = (gh)(c) = g(h(c)) = g(0) = 0$, which shows that $b = h(0) = 0$.

As TPH's post shows, we have $b = f(a) + h(c)$ for some pair $(a,c) \in A \times C$, for any element $b \in B$.

Suppose $b = f(a) + h(c) = f(a') + h(c')$.

Then $0 = f(a-a') + h(c-c')$. Hence, in $C$:

$0 = g(0) = (gf)(a-a') + (gh)(c-c')$

Since $gf = 0$ (by exactness, $g$ annihilates the entire image of $f$), this becomes:

$(gh)(c - c') = c - c' = 0$ that is: $c = c'$.

Thus:

$0 = f(a-a') + h(c-c') = f(a-a') + h(0) = f(a-a') + 0 = f(a-a')$.

Since $f$ is injective, and $a-a' \in \text{ker }f$, we have: $a - a' = 0$, so

$a = a'$, which shows that we have a UNIQUE pair, $(a,c)$.

The pay-off for this, is that if $D$ is ANY OTHER $R$-module, together with two $R$-module homomorphisms:

$f':A \to D$
$h':C \to D$

we can define a UNIQUE $R$-module homomorphism $k:B \to D$ by:

$k(b) = f'(a) + h'(c)$

and it is clear that:

$f' = kf, h' = kh$, since:

$f'(a) = k(f(a) + h(0)) = (kf)(a) + (kh)(0) = (kf)(a) + 0 = (kf)(a)$

and similarly for $h'$ (do you see "where" I have used the uniqueness of the pair $(a,c)$?).

But this is "the defining universal property" of the direct sum, which shows that $B$ must be isomorphic to $A \oplus C$ (up to a unique isomorphism).

 

[ThePerfectHacker]

"We say that the short-exact sequence,
$$ 0 \to A \to_f B \to_g C \to 0 $$
Is split iff there is a $h:C\to B$ such that $hg = 1$. "
Isn't the condition in this proposition $$ gh = 1 $$?

Yes, it is, I made the mistake of writing it backwards. But I think everything else still works through if you go through that paragraph again.

Another issue I am concerned about is that we say in (ii) above that we require $B_1 = f(A)$ and $g(B_2) = C$ with $g$ being an isomorphism between $B_2$ and $C$ ... ... ... but ... we do not seem anywhere to have shown that the homomorphism $g$ is an isomorphism between $B_2$ and $C$ ... ...

Yes, we never shown above that $g$ is an isomorphism between $B_2$ and $C$. But it is not hard to show. Recall the definition $B_2 = h(C)$. Now given $x\in B_2$ it means $x = h(y)$ for some $y\in C$, so $g(x) = g(h(y)) = y$. Therefore, if $x_1\not = x_2$ it means $x_1 = h(y_1)$ and $x_2=h(y_2)$ with $y_1\not = y_2$, applying $g$ yields $g(x_1) = y_1$ and $g(x_2) = y_2$, so $g(x_1)\not = g(x_2)$. This shows that the map $g:B_2\to C$ is injective. Can you show why it is surjective?

A third issue ... where I may be misunderstanding what is going on ... is as follows ... in your post you write:

" ... ... define $$ b_2 = h(g(b)) $$ ... ... "

... but since hg = 1 we have $$ h(g(b)) = b $$ ... but then this does not support us writing "set $$ b_1 = b - b_2 $$ ... can you clarify?

That was the mistake I wrote before, if you change it to $gh = 1$ then it is fine, so it should say $b_2 = h(g(b))$, but except $gh = 1$.