Exercise 2.19

Proof. For all $i\le j$, we have the commutative diagram

$$$$

with exact rows. So suppose $m\in M$; it can be written as ${\mu }_j(m_j)$ for some $m_j\in M_j$ by Exercise 2.15. Then,

by the commutativity of the diagram, hence $\mathrm{im}<!--FUNCTION\; APPLICATION-->\phi \subset \mathrm{Ker}<!--FUNCTION\; APPLICATION-->\psi$.

Conversely, suppose $n\in \mathrm{Ker}<!--FUNCTION\; APPLICATION-->\psi$, i.e., $\psi (n)=0$. By Exercise 2.15, write $n={\nu }_i(n_i)$ for some $n_i\in N_i$. Then, $\psi ({\nu }_i(n_i))={\xi }_i({\psi }_i(n_i))=0$. By the second part of Exercise 2.15, there exists $j\ge i$ such that ${\xi }_{ij}({\psi }_i(n_i))=0$ in $P_j$, hence by the commutativity of the diagram, ${\psi }_j({\nu }_{ij}(n_i))=0$. By exactness of the row $M_j\to N_j\to P_j$, this implies there exists $m_j\in M_j$ such that ${\phi }_j(m_j)={\nu }_{ij}(n_i)$. So, we have

by the commutativity of the diagram, hence $\mathrm{Ker}<!--FUNCTION\; APPLICATION-->\psi \subset \mathrm{im}<!--FUNCTION\; APPLICATION-->\phi$, and the bottom sequence $M\to N\to P$ is exact. □