The Four Lemma and Five Lemma

The four lemma and five lemma are similar to the snake lemma in the sense that they relate information between two exact sequences. However, these lemmas say something about the maps connecting the two sequences such as injectivity, surjectivity, or even when a map is an isomorphism. Most of the time these lemmas are used when the modules on the far left or far right in each sequence are both zero and the map connecting them is the identity map so as to trivially satisfy the assumptions.

Lemma 1. (Four Lemma)

1. Suppose that one has the following commutative diagram with exact rows: $\require{AMScd}$ $$\begin{CD} A @>{\varphi}>> B @>{\psi}>> C @>{\sigma}>> D\\ @VV{\alpha}V & @VV{\beta}V & @VV{\gamma}V & @VV{\delta}V\\ A' @>{\varphi'}>> B' @>{\psi'}>> C' @>{\sigma'}>> D \end{CD}$$ If $\beta$ and $\delta$ are injective and $\alpha$ is surjective, then $\gamma$ is injective.


2. Suppose that one has the following commutative diagram with exact rows: $$\begin{CD} B @>{\psi}>> C @>{\sigma}>> D@>{\tau}>> E\\ @VV{\beta}V & @VV{\gamma}V & @VV{\delta}V & @VV{\varepsilon}V\\ B' @>{\psi'}>> C' @>{\sigma'}>> D'@>{\tau'}>> E' \end{CD}$$ If $\beta$ and $\delta$ are surjective and $\varepsilon$ is injective, then $\gamma$ is surjective.

1. In order to show that $\gamma$ is injective, we will show that $\ker \gamma =0$. Let $x\in C$ be an element so that $\gamma(x) =0$. Then certainly the composition $\sigma'\big(\gamma(x)\big) =0$. However, by the commutativity of the diagram, $\delta\big(\sigma(x)\big) = \sigma'\big(\gamma(x)\big) =0$. However, $\delta$ is injective so it must be that $\sigma(x)=0$. Hence $x\in \ker \sigma = \im \psi$ by exactness, so there exists an element $y\in B$ so that $\psi(y) =x$. Now by commutativity of the diagram, we observe that $\psi'\big(\beta(y)\big) = \gamma\big(\psi(y)\big) = \gamma(x) =0$. Hence $\beta(y) \in \ker \psi' = \im \varphi'$, so there exists an element $z \in A'$ so that $\varphi'(z)= \beta(y)$. Now since $\alpha$ is surjective, there exists an element $a\in A$ so that $\alpha(a) = z$. Then by commutativity, we have that $$\beta\big(\varphi(a)\big)=\varphi'\big(\alpha(a)\big)=\varphi'(z) =\beta(y)$$ However, $\beta$ is injective so it must be that $y=\varphi(a)$. Finally, we see that $x= \psi(y) = \psi\big( \varphi(a)\big) =0$ due to exactness since $\psi\circ \varphi=0$. Hence $x=0$ so we conclude that $\ker \gamma =0$ so $\gamma$ is indeed injective.


2. Let $x\in C'$ be any element. Then since $\delta$ is a surjective map, there exists an element $y\in D$ so that $\delta(y) = \sigma'(x)$. By commutativity, $\tau'\big(\delta(y)\big) = \varepsilon\big(\tau(y)\big)$. Now since $\delta(y) = \sigma'(x)$ and exactness, we observe that $\varepsilon\big(\tau(y)\big) = \tau'\big(\delta(y)\big) =\tau'\big( \sigma'(x)\big) =0$ since $\tau'\circ \sigma' =0$. Hence $\tau(y) \in \ker \varepsilon = 0$ since $\varepsilon$ is injective. Hence $\tau(y) =0$ so $y\in \ker \tau = \im \sigma$ so there exists an element $z\in C$ so that $\sigma(z) = y$. Then by the commutativity of the diagram, we have that $\sigma'\big(\gamma(z)\big) = \delta\big(\sigma(z)\big) = \delta(y) = \sigma'(x)$. Now by linearity, we see that $\sigma'\big(x-\gamma(z)\big) =0$. Hence $x-\gamma(z)\in \ker \sigma' = \im \psi'$, hence there exists an element $w\in B'$ so that $\psi'(w) = x-\gamma(z)$. Now as $\beta$ is surjective, there exists an element $b\in B$ so that $\beta(b) = w$. Then by commutativity, $$\gamma\big(\psi(b)\big) = \psi'\big(\beta(b)\big) = \psi'(w) = x- \gamma(z)$$ Now by linearity, we see that $x= \gamma\big(\psi(b)\big) + \gamma(z) = \gamma\big(\psi(b)+z\big)$. Hence $x\in \im \gamma$ so we conclude that $\gamma$ is surjective.

Lemma 2. (Five Lemma)