# A question about split short exact sequence of modules

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, $\Bbb{Z}_2\oplus \Bbb{Z}_2\cong \langle (1,0)\rangle\oplus \langle (0,1)\rangle\cong \langle (1,0)\rangle\oplus \langle (1,1)\rangle$.)

• What is $M$ supposed to be? Jan 23, 2017 at 19:49
• 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. Jan 23, 2017 at 19:50
• Thanks for these recommendations. Jan 23, 2017 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 $C\cong A\oplus B$. Jan 23, 2017 at 20:01
• See here for an example. Jan 24, 2017 at 5:49

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 $C = \operatorname{Im} f \oplus \operatorname{Im} \beta$.

• Thanks. You teach me more than I asked. Jan 25, 2017 at 6:51
• 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 $\exists b\in B, \beta_1(b)\in \text{Im }\beta_1\backslash \text{Im }\beta_2 \Rightarrow \text{Im }\beta_1\backslash \text{Im }\beta_2\neq \emptyset \Rightarrow \text{Im }\beta_1\neq \text{Im }\beta_2$. Feb 19, 2017 at 3:45