The Stacks project

Proof. Suppose that the map $A^{\oplus n} \to A^{\oplus m}$ is given by the $m \times n$-matrix $(a_{ij})$. Consider the ring $B = A[x_1, \ldots , x_ n]/(\sum a_{ij}x_ j)$. The element $(x_1, \ldots , x_ n) \in \underline{A^{\oplus n}}(B)$ maps to zero in $\underline{A^{\oplus m}}(B)$ hence is the image of a unique element $\xi \in L(B)$. Note that $\xi $ has the following universal property: for any $A$-algebra $C$ and any $\xi ' \in L(C)$ there exists an $A$-algebra map $B \to C$ such that $\xi $ maps to $\xi '$ via the map $L(B) \to L(C)$.

Note that $B$ is a graded $A$-algebra, hence we can use Lemmas 46.3.7 and 46.3.5 to decompose the values of our functors on $B$ into graded pieces. Note that $\xi \in L(B)^{(1)}$ as $(x_1, \ldots , x_ n)$ is an element of degree one in $\underline{A^{\oplus n}}(B)$. Hence we see that $\varphi (\xi ) \in \underline{M}(B)^{(1)} = M \otimes _ A B_1$. Since $B_1$ is generated by $x_1, \ldots , x_ n$ as an $A$-module we can write $\varphi (\xi ) = \sum m_ i \otimes x_ i$. Consider the map $A^{\oplus n} \to M$ which maps the $i$th basis vector to $m_ i$. By construction the associated map $\underline{A^{\oplus n}} \to \underline{M}$ maps the element $\xi $ to $\varphi (\xi )$. It follows from the universal property mentioned above that the diagram commutes. $\square$