PlanetMath: horseshoe lemma

Let $\mathcal{A}$ be an abelian category with enough projectives. The horseshoe lemma, also called the simultaneous resolution theorem, says that if

Proof. We fill in the diagram a column at a time, proving exactness along the way. First we construct a surjective map $\pi\colon P_0\to A$ . There is a map from

, the composition $P'_0\to A'\to A$ . Since

is projective, there is a filler for the diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ & P''_0\ar[d]\ar@{.>}[ld] & \ A\ar[r] & A''\ar[r] & 0 } } \end{xy}$

Since

is the coproduct of

and

, there is a unique filler $\pi$ for the diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ P'_0\ar[r]\ar[rd] & P_0\ar@{.>}[d]^{\pi} & P''_0\ar[l]\ar[ld] \ & A & } } \end{xy}$

The diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ 0\ar[r] & P'_0\ar[r]\ar[d] & P_0... ...0\ar[r]\ar[d] & 0 \ 0\ar[r] & A'\ar[r] & A\ar[r] & A''\ar[r] & 0 } } \end{xy}$

is commutative and has exact rows, so we may apply the short five lemma to conclude that $\pi$ is surjective.

Now assume for induction that we have constructed a partial resolution of in this manner, yielding a diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ 0\ar[d] & 0\ar[d] & & 0\ar[d] & ... ...ar[r]\ar[d] & \cdots\ar[r] & A''\ar[r]\ar[d] & 0 \ 0 & 0 & & 0 & } } \end{xy}$

with exact rows and columns. (We are not assuming here that

; if

, then $P'_{n-1}$ denotes

and $P'_{n-2}$ denotes 0; similar substitutions apply). By the snake lemma, there is a commutative diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ 0\ar[r] & \ker d'_n\ar[r]\ar[d] ... ...\ 0\ar[r] & P'_{n-2}\ar[r] & P_{n-2}\ar[r] & P''_{n-2}\ar[r] & 0 } } \end{xy}$

with exact rows and columns. In fact, the map $\ker d_n\to\ker d''_n$ is surjective; this can be verified by a diagram chase. We can construct a surjective map $b_{n+1}\colon P_{n+1}\to\ker d_n$ in the same way we constructed $\pi$ . Specifically, there is a map $P'_{n+1}\to\ker d'_n$ obtained by composition, and there is a filler for the diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ & P''_n\ar[d]\ar@{.>}[ld] & \ \ker d_n\ar[r] & \ker d''_n\ar[r] & 0 } } \end{xy}$

These maps determine $b_{n+1}$ uniquely. Define $d_{n+1}\colon P_{n+1}\to P_n$ as the composition $P_{n+1}\xrightarrow{b_{n+1}} \ker d_n\hookrightarrow P_n$ ; since $b_{n+1}$ is surjective, the sequence $P_{n+1}\to P_n\to P_{n-1}$ is exact. Hence we have managed to construct a diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ 0\ar[d] & 0\ar[d] & 0\ar[d] & & ... ...]\ar[d] & \cdots\ar[r] & A''\ar[r]\ar[d] & 0 \ 0 & 0 & 0 & & 0 & } } \end{xy}$

with exact rows and columns. This completes the proof. $\qedsymbol$

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