Five Lemma

Theorem

Let $A$ be a commutative ring with unity.

Let:

$\begin{xy}\xymatrix@L+2mu@+1em{ M_1 \ar[r]^*{\alpha_1} \ar[d]^*{\phi_1} & M_2 \ar[r]^*{\alpha_2} \ar[d]^*{\phi_2} & M_3 \ar[r]^*{\alpha_3} \ar[d]^*{\phi_3} & M_4 \ar[r]^*{\alpha_4} \ar[d]^*{\phi_4} & M_5 \ar[d]^*{\phi_5} \\ N_1 \ar[r]_*{\beta_1} & N_2 \ar[r]_*{\beta_2} & N_3 \ar[r]_*{\beta_3} & N_4 \ar[r]_*{\beta_4} & N_5 }\end{xy}$

be a commutative diagram of $A$-modules.

Suppose that the rows are exact.

Then:

If $\phi_2$ and $\phi_4$ are surjective and $\phi_5$ is injective then $\phi_3$ is surjective.

If $\phi_2$ and $\phi_4$ are injective and $\phi_1$ is surjective then $\phi_3$ is injective.

Proof

First suppose that $\phi_2$ and $\phi_4$ are surjective and $\phi_5$ is injective.

Let $n_3 \in N_3$ be any element.

We want to find $x \in M_3$ such that $\map {\phi_3} x = n_3$.

Let $n_4 = \map {\beta_3} {n_3} \in N_4$.

Since $\phi_4$ is surjective, there exists $m_4 \in M_4$ such that $\map {\phi_4} {m_4} = n_4$.

Since the rows are exact, we have that:

$\map {\beta_4} {n_4} = \map {\beta_4 \circ \beta_3} {n_3} = 0$

That is:

$\map {\beta_4 \circ \phi_4} {m_4} = 0$

Since the diagram is commutative, this implies that $\map {\phi_5 \circ \alpha_4} {m_4} = 0$.

Since $\phi_5$ is injective, this means that $\map {\alpha_4} {m_4} = 0$.

That is:

$m_4 \in \ker \alpha_4$

Since the rows are exact, this means that $m_4 \in \Img {\alpha_3}$, say:

$m_4 = \map {\alpha_3} {m_3}$

Thus we have:

$\map {\beta_3} {n_3} = n_4 = \map {\phi_4 \circ \alpha_3} {m_3} = \map {\beta_3 \circ \phi_3} {m_3}$

Using the fact that $\beta_3$ is a homomorphism this means that:

$n_3 - \map {\phi_3} {m_3} \in \ker \beta_3$

By exactness of the rows we therefore have $n_3 - \map {\phi_3} {m_3} = \map {\beta_2} {n_2}$ for some $n_2 \in N_2$.

By hypothesis $\phi_2$ is surjective, so there exists $m_2 \in M_2$ such that:

$n_3 - \map {\phi_3} {m_3} = \map {\beta_2 \circ \phi_2} {m_2}$

Since the diagram is commutative, we have therefore:

$n_3 - \map {\phi_3} {m_3} = \map {\phi_3 \circ \alpha_2} {m_2}$

Finally by the homomorphism property this shows that:

$n_3 = \map {\phi_3} {m_3 + \map {\alpha_2} {m_2} }$

as required.

Now suppose that $\phi_2$ and $\phi_4$ are injective and $\phi_1$ is surjective.

Let $m_3 \in M_3$ such that $\map {\phi_3} {m_3} = 0$.

We want to show that $m_3 = 0$.

First of all $\map {\phi_3} {m_3} = 0$ implies that $\map {\beta_3 \circ \phi_3} {m_3} = 0$.

Since the diagram is commutative, this implies that $\map {\phi_4 \circ \alpha_3} {m_3} = 0$.

Since $\phi_4$ is injective we have:

$\map {\alpha_3} {m_3} = 0$

Since the rows are exact, we therefore have $m_3 = \map {\alpha_2} {m_2}$ for some $m_2 \in M_2$.

Let $n_2 = \map {\phi_2} {m_2} \in N_2$.

By commutativity of the diagram, we have:

$\map {\beta_2} {n_2} = \map {\phi_3 \circ \alpha_2} {m_2} = \map {\phi_3} {m_3} = 0$

By exactness of the rows it follows that there exists $n_1 \in N_1$ such that $n_2 = \map {\beta_1} {n_1}$.

Since $\phi_1$ is assumed surjective this means that $n_1 = \map {\phi_1} {m_1}$ for some $m_1 \in M_1$.

Now:

$\map {\phi_2} {m_2} = \map {\beta_1 \circ \phi_1} {m_1} = \map {\phi_2 \circ \alpha_1} {m_1}$

Thus by the homomorphism property:

$m_2 - \map {\alpha_1} {m_1} \in \ker \phi_2$

Since $\phi_2$ is assumed injective this means that $m_2 = \map {\alpha_1} {m_1}$.

Finally, since the rows are exact we have:

$m_3 = \map {\alpha_2} {m_2} = \map {\alpha_2 \circ \alpha_1} {m_1} = 0$

as required.

$\blacksquare$