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Let $0\longrightarrow A\stackrel{f}{\longrightarrow} C\stackrel{g}{\longrightarrow} B\longrightarrow 0$ be a split short exact sequence of modules.

That is, there exists $\alpha:C\to A$ such that $\alpha\circ f=1_{A}$ or there exists $\beta:B\to C$ such that $g\circ \beta=1_{B}$.

We have \begin{eqnarray*} C &\cong& \text{Im }f\oplus \ker{\alpha}\\ &\cong& \ker{g}\oplus \text{Im }\beta. \end{eqnarray*} It seems $\ker{\alpha}=\text{Im }\beta$ because $\text{Im }f=\ker{g}$. But I can't prove it and can't find a counterexample. Is it ($\ker{\alpha}=\text{Im }\beta$) true?

(There is an example in group theory which shows that $A\oplus B\cong A\oplus C$ doesn't imply $B=C$. For example, $\Bbb{Z}_2\oplus \Bbb{Z}_2\cong \langle (1,0)\rangle\oplus \langle (0,1)\rangle\cong \langle (1,0)\rangle\oplus \langle (1,1)\rangle$.)

bfhaha
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  • What is $M$ supposed to be? – Jack Lee Jan 23 '17 at 19:49
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    From the tags I would guess that you are talking about short exact sequences of modules, but this is worth clarifying as there are other kinds of short exact sequences. – Spenser Jan 23 '17 at 19:50
  • Thanks for these recommendations. – bfhaha Jan 23 '17 at 19:53
  • It is important that the fact that the short exact sequence is split is not equivalent to the existence of an isomorphism $C\cong A\oplus B$. – Mariano Suárez-Álvarez Jan 23 '17 at 20:01
  • Theorem IV.1.18 in Hungerford's Algebra: Let $R$ be a rin and $0\to A_1\stackrel{f}{\to}B\stackrel{g}{\to}A_2\to 0$ a short exact sequence of $R$-module homomorphisms. Then the following conditions are equivalent. (i) There is an $R$-module homomorphism $h:A_2\to B$ with $gh=1_{A_2}$; (ii) Ther is an $R$-module homomorphism $k:B\to A_1$ with $kf=1_{A_1}$; (iii) the given sequence is isomorphic (with identity maps on $A_1$ and $A_2$) to the direct sum short exact sequence $0\to A_1\stackrel{\iota_1}{\to}A_1\oplus A_2\stackrel{\pi_2}{\to}A_2\to 0$; in particular $B\cong A_1\oplus A_2$. – bfhaha Jan 24 '17 at 05:24
  • @MarianoSuárez-Álvarez Really? But my textbook states that:

    A short exact sequence that satisfies the equivalent conditions of Theorem IV.1.18 is said to be split or a split exact sequence.

     – bfhaha Jan 24 '17 at 05:24
  • The condition (III) in that statement is considerably stronger than the mere existence of an isomorphism between the middle term and the direct sum of the other two: it says that the given sequence is isomorphic as a sequence to the one with the direct sum in the middle and the specific maps described there. And with good reason: as I said, there are non split short exact sequences in which the middle term is the direct sum of the other two. – Mariano Suárez-Álvarez Jan 24 '17 at 05:46
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    See here for an example. – Mariano Suárez-Álvarez Jan 24 '17 at 05:49
  • @MarianoSuárez-Álvarez Thanks! I will clarify the misunderstanding. – bfhaha Jan 24 '17 at 05:53

1 Answers1

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Note that if the sequence splits, then there are typically many sections of $g$. If $\beta \colon B \to C$ satisfies $g\circ \beta = 1_B$, and $\delta \colon B \to \ker g$ is any homomorphism, then $\beta + \delta$ is also a section of $g$:

$$g\bigl((\beta + \delta)(b)\bigr) = g\bigl(\beta(b) + \delta(b)\bigr) = g\bigl(\beta(b)\bigr) + g\bigl(\underbrace{\delta(b)}_{\in \ker g}\bigr) = g\bigl(\beta(b)\bigr) = b.$$

Conversely, if $\beta_1,\beta_2$ are sections of $g$, then $\operatorname{Im} (\beta_1 - \beta_2) \subset \ker g$, so the above construction yields all sections of $g$.

And any two different sections of $g$ have different image. If $\beta_1,\beta_2$ are different sections of $g$, and $b\in B$ is such that $\beta_1(b) \neq \beta_2(b)$, then $\beta_1(b) \in \operatorname{Im} \beta_1 \setminus \operatorname{Im} \beta_2$. For if there were $b'\in B$ with $\beta_1(b) = \beta_2(b')$, then $b = g\bigl(\beta_1(b)\bigr) = g\bigl(\beta_2(b')\bigr) = b'$, contradicting $\beta_1(b) \neq \beta_2(b)$.

Thus in general, we cannot expect $\ker \alpha = \operatorname{Im} \beta$.

For a concrete example, let $R$ be a (nontrivial) ring, and consider

$$0 \to R \xrightarrow{r \mapsto (r,0)} R \times R \xrightarrow{(r,s) \mapsto s} R \to 0$$

with $\alpha(r,s) = r$, so $\ker \alpha = \{0\} \times R$ and $\beta(s) = (s,s)$. The image of $\beta$ is the diagonal of $R\times R$, and $\ker \alpha \cap \operatorname{Im} \beta = \{(0,0)\}$.

However, if the sequence splits, then we can always choose $\alpha,\beta$ in such a way that $\ker \alpha = \operatorname{Im} \beta$.

If $\alpha$ is given, then

$$\bigl(g\lvert_{\ker \alpha}\bigr) \colon \ker \alpha \to B$$

is a module isomorphism, and we can choose $\beta = \bigl(g\lvert_{\ker \alpha}\bigr)^{-1}$ [and of course this is the only section of $g$ with image $\ker \alpha$].

If $\beta$ is given, then $p = \beta \circ g \colon C \to C$ is a projection, $p \circ p = \beta \circ (g\circ \beta)\circ g = \beta \circ 1_B \circ g = \beta \circ g = p$, and $\ker p = \ker g = \operatorname{Im} f$, so $\alpha = f^{-1}\circ (1_C - p)$ is a homomorphism that satisfies $\alpha \circ f = 1_A$ and $\ker \alpha = \ker (1_C - p) = \operatorname{Im} p = \operatorname{Im} \beta$. Of course $\alpha$ is uniquely determined by the conditions $\alpha \circ f = 1_A$ and $\ker \alpha = \operatorname{Im} \beta$, since $C = \operatorname{Im} f \oplus \operatorname{Im} \beta$.

Daniel Fischer
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    Thanks. You teach me more than I asked. – bfhaha Jan 25 '17 at 06:51
  • Just a note for myself. $\beta_1\neq \beta_2 \Rightarrow \exists b\in B, \beta_1(b)\neq \beta_2(b)$. On the other hand, $\forall b'\neq b\in B, g(\beta_1(b))=b\neq b'=g(\beta_2(b')) \Rightarrow \forall b'\neq b\in B, \beta_1(b)\neq \beta_2(b')$. Thus, $\forall x\in B, \beta_1(b)\neq \beta_2(x)$. It follows that $\exists b\in B, \beta_1(b)\in \text{Im }\beta_1\backslash \text{Im }\beta_2 \Rightarrow \text{Im }\beta_1\backslash \text{Im }\beta_2\neq \emptyset \Rightarrow \text{Im }\beta_1\neq \text{Im }\beta_2$. – bfhaha Feb 19 '17 at 03:45