The Snake Lemma for modules

The restrictions $\ker a\to \ker b$ and $\ker b\to \ker c$ are induced by $i$ and $p$ using [C1] and [C2]. The formulas $[x^{\mathrm{\prime }}]\mapsto [i^{\mathrm{\prime }}(x^{\mathrm{\prime }})]$ and $[y^{\mathrm{\prime }}]\mapsto [p^{\mathrm{\prime }}(y^{\mathrm{\prime }})]$ define maps $\mathrm{coker}a\to \mathrm{coker}b$ and $\mathrm{coker}b\to \mathrm{coker}c$: [C1] and [C2] make the relevant images vanish in the target quotients, so [L1] applies.

For $z\in \ker c$, choose $y\in B$ with $p(y)=z$. Then [C2] gives $p^{\mathrm{\prime }}(b(y))=c(z)=0$, so [F1] gives a unique $x^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$ with $i^{\mathrm{\prime }}(x^{\mathrm{\prime }})=b(y)$. Define $\delta (z):=[x^{\mathrm{\prime }}]\in \mathrm{coker}a$.

If $\tilde{y}$ is another lift of $z$, then $\tilde{y}-y=i(x)$ for some $x\in A$ by [F1]. If $i^{\mathrm{\prime }}({\tilde{x}}^{\mathrm{\prime }})=b(\tilde{y})$, then [C1] and injectivity of $i^{\mathrm{\prime }}$ give ${\tilde{x}}^{\mathrm{\prime }}-x^{\mathrm{\prime }}=a(x)$, so $[{\tilde{x}}^{\mathrm{\prime }}]=[x^{\mathrm{\prime }}]$ in $\mathrm{coker}a$. Thus $\delta$ is well defined.

Exactness at $\ker a$ holds because its map is the restriction of the injective map $i$. At $\ker b$, the composite induced by $p\circ i$ is zero; if $y\in \ker b$ maps to zero in $\ker c$, then $p(y)=0$, so $y=i(x)$ by [F1], and [C1] with injectivity of $i^{\mathrm{\prime }}$ gives $a(x)=0$, hence $x\in \ker a$.

If $y\in \ker b$, the construction of step 1.2 applied to $z=p(y)$ has $x^{\mathrm{\prime }}=0$, so $\delta (z)=0$. Conversely, if $z\in \ker c$ has $\delta (z)=0$, choose $y,x^{\mathrm{\prime }}$ as in step 1.2; then $x^{\mathrm{\prime }}=a(x)$ for some $x$, so [C1] gives $b(y-i(x))=0$ and $p(y-i(x))=z$. Thus exactness holds at $\ker c$.

The map $\mathrm{coker}a\to \mathrm{coker}b$ kills $\delta (z)$ because $i^{\mathrm{\prime }}(x^{\mathrm{\prime }})=b(y)$. Conversely, if $[x^{\mathrm{\prime }}]$ maps to zero, write $i^{\mathrm{\prime }}(x^{\mathrm{\prime }})=b(y)$; then [C2] gives $c(p(y))=0$, and the construction with lift $y$ gives $\delta (p(y))=[x^{\mathrm{\prime }}]$. Thus exactness holds at $\mathrm{coker}a$.

The next composite is zero because $p^{\mathrm{\prime }}{i}^{\mathrm{\prime }}=0$. If $[y^{\mathrm{\prime }}]\in \mathrm{coker}b$ maps to zero in $\mathrm{coker}c$, write $p^{\mathrm{\prime }}(y^{\mathrm{\prime }})=c(z)$, choose $y\in B$ with $p(y)=z$, and use [C2] to obtain $y^{\mathrm{\prime }}-b(y)\in \ker p^{\mathrm{\prime }}=\mathrm{im}{i}^{\mathrm{\prime }}$. Hence $[y^{\mathrm{\prime }}]$ comes from $\mathrm{coker}a$, proving exactness at $\mathrm{coker}b$.

The map $\mathrm{coker}b\to \mathrm{coker}c$ is surjective: for a class $[z^{\mathrm{\prime }}]$, choose $y^{\mathrm{\prime }}\in B^{\mathrm{\prime }}$ with $p^{\mathrm{\prime }}(y^{\mathrm{\prime }})=z^{\mathrm{\prime }}$ using [F1], and $[y^{\mathrm{\prime }}]$ maps to $[z^{\mathrm{\prime }}]$.

For $z_1,z_2\in \ker c$ and $r\in R$, the choices $y_1+y_2$ and $r{y}_1$ in step 1.2 lead to $x_1^{\mathrm{\prime }}+x_2^{\mathrm{\prime }}$ and $r{x}_1^{\mathrm{\prime }}$; uniqueness through the injective map $i^{\mathrm{\prime }}$ then gives $\delta (z_1+z_2)=\delta (z_1)+\delta (z_2)$ and $\delta (r{z}_1)=r\delta (z_1)$.

Steps 2.2 through 2.6 establish exactness at every displayed term. Steps 1.2, 2.1, and 3.1 construct a well-defined linear connecting homomorphism.