**Proof.**
Obviously (1) implies (2).

Next we show (2) implies (3). Let $f_1(x_ i) = \sum _ j a_{ij} y_ j$ be relations as in (3). Let $(d_ j)$ be a basis for $R^ m$, $(e_ i)$ a basis for $R^ n$, and $R^ m \to R^ n$ the map given by $d_ j \mapsto \sum _ i a_{ij} e_ i$. Let $Q$ be the cokernel of $R^ m \to R^ n$. Then tensoring $R^ m \to R^ n \to Q \to 0$ by the map $f_1: M_1 \to M_2$, we get a commutative diagram

\[ \xymatrix{ M_1^{\oplus m} \ar[r] \ar[d] & M_1^{\oplus n} \ar[r] \ar[d] & M_1 \otimes _ R Q \ar[r] \ar[d] & 0 \\ M_2^{\oplus m} \ar[r] & M_2^{\oplus n} \ar[r] & M_2 \otimes _ R Q \ar[r] & 0 } \]

where $M_1^{\oplus m} \to M_1^{\oplus n}$ is given by

\[ (z_1, \ldots , z_ m) \mapsto (\sum \nolimits _ j a_{1j} z_ j, \ldots , \sum \nolimits _ j a_{nj} z_ j), \]

and $M_2^{\oplus m} \to M_2^{\oplus n}$ is given similarly. We want to show $x = (x_1, \ldots , x_ n) \in M_1^{\oplus n}$ is in the image of $M_1^{\oplus m} \to M_1^{\oplus n}$. By (2) the map $M_1 \otimes Q \to M_2 \otimes Q$ is injective, hence by exactness of the top row it is enough to show $x$ maps to $0$ in $M_2 \otimes Q$, and so by exactness of the bottom row it is enough to show the image of $x$ in $M_2^{\oplus n}$ is in the image of $M_2^{\oplus m} \to M_2^{\oplus n}$. This is true by assumption.

Condition (4) is just a translation of (3) into diagram form.

Next we show (4) implies (5). Let $\varphi : P \to M_3$ be a map from a finitely presented $R$-module $P$. We must show that $\varphi $ lifts to a map $P \to M_2$. Choose a presentation of $P$,

\[ R^ n \xrightarrow {g_1} R^ m \xrightarrow {g_2} P \to 0. \]

Using freeness of $R^ n$ and $R^ m$, we can construct $h_2: R^ m \to M_2$ and then $h_1: R^ n \to M_1$ such that the following diagram commutes

\[ \xymatrix{ & R^ n \ar[r]^{g_1} \ar[d]^{h_1} & R^ m \ar[r]^{g_2} \ar[d]^{h_2} & P \ar[r] \ar[d]^{\varphi } & 0 \\ 0 \ar[r] & M_1 \ar[r]^{f_1} & M_2 \ar[r]^{f_2} & M_3 \ar[r] & 0 . } \]

By (4) there is a map $k_1: R^ m \to M_1$ such that $k_1 \circ g_1 = h_1$. Now define $h'_2: R^ m \to M_2$ by $h_2' = h_2 - f_1 \circ k_1$. Then

\[ h'_2 \circ g_1 = h_2 \circ g_1 - f_1 \circ k_1 \circ g_1 = h_2 \circ g_1 - f_1 \circ h_1 = 0 . \]

Hence by passing to the quotient $h'_2$ defines a map $\varphi ': P \to M_2$ such that $\varphi ' \circ g_2 = h_2'$. In a diagram, we have

\[ \xymatrix{ R^ m \ar[r]^{g_2} \ar[d]_{h'_2} & P \ar[d]^{\varphi } \ar[dl]_{\varphi '} \\ M_2 \ar[r]^{f_2} & M_3. } \]

where the top triangle commutes. We claim that $\varphi '$ is the desired lift, i.e. that $f_2 \circ \varphi ' = \varphi $. From the definitions we have

\[ f_2 \circ \varphi ' \circ g_2 = f_2 \circ h'_2 = f_2 \circ h_2 - f_2 \circ f_1 \circ k_1 = f_2 \circ h_2 = \varphi \circ g_2. \]

Since $g_2$ is surjective, this finishes the proof.

Now we show (5) implies (6). Write $M_{3}$ as the colimit of a directed system of finitely presented modules $M_{3, i}$, see Lemma 10.11.3. Let $M_{2, i}$ be the fiber product of $M_{3, i}$ and $M_{2}$ over $M_{3}$—by definition this is the submodule of $M_2 \times M_{3, i}$ consisting of elements whose two projections onto $M_3$ are equal. Let $M_{1, i}$ be the kernel of the projection $M_{2, i} \to M_{3, i}$. Then we have a directed system of exact sequences

\[ 0 \to M_{1, i} \to M_{2, i} \to M_{3, i} \to 0, \]

and for each $i$ a map of exact sequences

\[ \xymatrix{ 0 \ar[r] & M_{1, i} \ar[d] \ar[r] & M_{2, i} \ar[r] \ar[d] & M_{3, i} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & M_{1} \ar[r] & M_{2} \ar[r] & M_{3} \ar[r] & 0 } \]

compatible with the directed system. From the definition of the fiber product $M_{2, i}$, it follows that the map $M_{1, i} \to M_1$ is an isomorphism. By (5) there is a map $M_{3, i} \to M_{2}$ lifting $M_{3, i} \to M_3$, and by the universal property of the fiber product this gives rise to a section of $M_{2, i} \to M_{3, i}$. Hence the sequences

\[ 0 \to M_{1, i} \to M_{2, i} \to M_{3, i} \to 0 \]

split. Passing to the colimit, we have a commutative diagram

\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{colim}}\nolimits M_{1, i} \ar[d]^{\cong } \ar[r] & \mathop{\mathrm{colim}}\nolimits M_{2, i} \ar[r] \ar[d] & \mathop{\mathrm{colim}}\nolimits M_{3, i} \ar[d]^{\cong } \ar[r] & 0 \\ 0 \ar[r] & M_{1} \ar[r] & M_{2} \ar[r] & M_{3} \ar[r] & 0 } \]

with exact rows and outer vertical maps isomorphisms. Hence $\mathop{\mathrm{colim}}\nolimits M_{2, i} \to M_2$ is also an isomorphism and (6) holds.

Condition (6) implies (1) by Example 10.82.2 (2).
$\square$

## Comments (4)

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