91Ó°ÊÓ

The problem statement already specifies some definitions and assumptions, but let's provide formal definitions for all the terms involved: - Injective module: A module \(I\) is injective if, for every module \(A\) and every submodule \(B \subset A\), whenever there is a homomorphism \(f: B \rightarrow I\), there exists a homomorphism \(\tilde{f}: A \rightarrow I\) extending \(f\). - Exact sequence: A sequence of homomorphisms between modules \( \cdots \rightarrow M_{i-1} \xrightarrow{\phi_{i-1}} M_i \xrightarrow{\phi_i} M_{i+1} \rightarrow \cdots\) is exact if the image of each homomorphism is equal to the kernel of the next homomorphism, i.e., \(\text{Im}(\phi_{i-1}) = \text{Ker}(\phi_i)\) for all \(i\). - Split sequence: An exact sequence \(0 \rightarrow A \xrightarrow{\alpha} B \xrightarrow{\beta} C \rightarrow 0\) is said to split if there exist homomorphisms \(\gamma: B \rightarrow A\) and \(\delta: C \rightarrow B\) such that \(\alpha \circ \gamma = \text{Id}_A\) and \(\delta \circ \beta = \text{Id}_C\). This means that \(\alpha\) is injective, \(\beta\) is surjective, and there exist homomorphisms \(\gamma\) and \(\delta\) that satisfy the conditions. Now that we have defined these concepts, we can start the proof of the three parts of the exercise.

Given that \(I_{1}\) and \(I_{2}\) are injective, we need to find homomorphisms \(\gamma: I_{2} \rightarrow I_{1}\) and \(\delta: I_{3} \rightarrow I_{2}\) such that the splitting conditions are satisfied. Consider \(f = \text{Id}_{I_{1}}: I_{1} \rightarrow I_{1}\). By the injectivity of \(I_{1}\), there exists an extension \(\tilde{f}: I_{2} \rightarrow I_{1}\) such that \(\tilde{f} \circ \text{Id}_{I_{1}} = \tilde{f}\). Define \(\gamma=\tilde{f}\). Then we have \(\gamma \circ \text{id}_{I_{1}}=\gamma\). It is clear that \(\gamma \circ I_{1} = I_{1} \), which implies \(\gamma \circ I_{1} = \text{Id}_{I_{1}}\). Now, let's consider \(g: \text{Ker}(I_3) \rightarrow I_{2}\). Since \(\text{Ker}(I_3)=\text{Im}(I_2)\), we can define \(g=\text{id}_{\text{Im}(I_2)}\). As \(I_2\) is injective, there must exist an extension \(\tilde{g}: I_{3} \rightarrow I_{2}\) such that \(\tilde{g} \circ I_{2} = I_{2}\). Define \(\delta=\tilde{g}\). Then we have \(\delta \circ \text{id}_{\text{Im}(I_2)}=\delta\). It follows that \(\delta \circ I_{3} = I_{3}\), which implies that \(\delta \circ I_{3} = \text{Id}_{I_{3}}\). The exact sequence \(0 \rightarrow I_{1} \rightarrow I_{2} \rightarrow I_{3} \rightarrow 0\) splits because we found the homomorphisms \(\gamma\) and \(\delta\) that satisfy the splitting conditions.

To prove that \(I_{3}\) is injective, we will show that for every module \(A\) and every submodule \(B\subset A\), if there is a homomorphism \(h: B \rightarrow I_{3}\), then there exists a homomorphism \(\tilde{h}: A \rightarrow I_{3}\) extending \(h\). First, consider the composition \(h \circ \delta: B \rightarrow I_{2}\). Since \(I_{2}\) is injective, there exists a homomorphism \(\tilde{g}: A \rightarrow I_{2}\) extending \(h \circ \delta\). Now, consider the composition \(\tilde{g} \circ I_{3}: A \rightarrow I_{3}\). Define \(\tilde{h} = \tilde{g} \circ I_{3}\). For the restriction of \(\tilde{h}\) to \(B\), we see that \((\tilde{h} \circ I_{3})|_B = \tilde{g} |_B = h\). Therefore, \(\tilde{h}\) is an extension of \(h\). Since this holds for any module \(A\) and submodule \(B\subset A\), we conclude that \(I_{3}\) is injective.

Let \(A\) be any module and \(B\subset A\) be a submodule. We need to show that if there is a homomorphism \(k: B \rightarrow M\), then there exists a homomorphism \(\tilde{k}: A \rightarrow M\) extending \(k\). Consider the homomorphism \(k': B \rightarrow M \oplus N\), defined as \(k'(b)=(k(b), 0)\). Since \(I = M \oplus N\), and \(I\) is injective, there exists a homomorphism \(\tilde{k'}: A \rightarrow M \oplus N\) extending \(k'\). Now, define a projection \(\pi_M: M \oplus N \rightarrow M\) as \(\pi_M(m, n)=m\). Then, define \(\tilde{k} = \pi_M \circ \tilde{k'}: A \rightarrow M\). We can now verify that \(\tilde{k}\) extends \(k\): For any \(b \in B\), we have \[ \tilde{k}(b) = \pi_M(\tilde{k'}(b)) = \pi_M(k'(b)) = \pi_M(k(b), 0) = k(b). \] Thus, we have found a homomorphism \(\tilde{k}: A \rightarrow M\) that extends \(k\). Since this holds for any module \(A\) and its submodule \(B \subset A\), we conclude that \(M\) is injective.