Let A\mathcal A be an . Consider a commutative 3×33\times 3 diagram

0AAA00BBB00CCC0\begin{array}{ccccccccc} 0 &\to& A' &\xrightarrow{}& A &\xrightarrow{}& A'' &\to& 0\\ && \downarrow && \downarrow && \downarrow && \\ 0 &\to& B' &\xrightarrow{}& B &\xrightarrow{}& B'' &\to& 0\\ && \downarrow && \downarrow && \downarrow && \\ 0 &\to& C' &\xrightarrow{}& C &\xrightarrow{}& C'' &\to& 0 \end{array}

If the three columns and the first two rows are , then the third row

0CCC00\to C' \to C \to C'' \to 0

is exact. Dually, exact rows together with two exact columns force the remaining column to be exact.

Remarks

The lemma can be proved by a diagram chase using the , or by a kernel–cokernel argument in the abelian category.