Lemma 12.5.19. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ w \ar[r] \ar[d]^\alpha & x \ar[r] \ar[d]^\beta & y \ar[r] \ar[d]^\gamma & z \ar[d]^\delta \\ w' \ar[r] & x' \ar[r] & y' \ar[r] & z' } \]

be a commutative diagram with exact rows.

If $\alpha , \gamma $ are surjective and $\delta $ is injective, then $\beta $ is surjective.

If $\beta , \delta $ are injective and $\alpha $ is surjective, then $\gamma $ is injective.

**Proof.**
Assume $\alpha , \gamma $ are surjective and $\delta $ is injective. We may replace $w'$ by $\mathop{\mathrm{Im}}(w' \to x')$, i.e., we may assume that $w' \to x'$ is injective. We may replace $z$ by $\mathop{\mathrm{Im}}(y \to z)$, i.e., we may assume that $y \to z$ is surjective. Then we may apply Lemma 12.5.17 to

\[ \xymatrix{ & \mathop{\mathrm{Ker}}(y \to z) \ar[r] \ar[d] & y \ar[r] \ar[d] & z \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathop{\mathrm{Ker}}(y' \to z') \ar[r] & y' \ar[r] & z' } \]

to conclude that $\mathop{\mathrm{Ker}}(y \to z) \to \mathop{\mathrm{Ker}}(y' \to z')$ is surjective. Finally, we apply Lemma 12.5.17 to

\[ \xymatrix{ & w \ar[r] \ar[d] & x \ar[r] \ar[d] & \mathop{\mathrm{Ker}}(y \to z) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & w' \ar[r] & x' \ar[r] & \mathop{\mathrm{Ker}}(y' \to z') } \]

to conclude that $x \to x'$ is surjective. This proves (1). The proof of (2) is dual to this.
$\square$

## Comments (2)

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