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Let $R$ be a commutative ring with identity and let $$\matrix{&&X\\&&\downarrow\\Y&\to&M}$$ be homomorphisms of $R$-modules. Show that it can be embedded into some pullback square $$\matrix{L&\to&X\\\downarrow&&\downarrow\\Y&\to&M}$$

I am having trouble constructing $L$. I tried using the same construction as proving the pushout square by setting $L=(X\oplus Y)/N$ for some suitable $N$, but this rendered $L\to X$ and $L\to Y$ not well-defined.

Any hints?

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    $\begingroup$ Do you know how pullbacks are defined for sets? $\endgroup$ – Arnaud D. Jun 13 at 13:12
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    $\begingroup$ Define $L$ as a submodule of $L\oplus Y$ rather than a quotient. $\endgroup$ – user10354138 Jun 13 at 13:19
  • $\begingroup$ Suppose $f:X\to M$ and $g:Y\to M$ are the given homomorphisms. Let $L=f^{-1}(f(X)\cap g(Y))\oplus g^{-1}(f(X)\cap g(Y))$. Does this make sense? $\endgroup$ – trisct Jun 13 at 13:22
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Let $+_A $ and $*_A$ denote the operations on an R-module A.

Let $L = \{(x,y): f(x) = g(y)\}$ where $f$ and $g$ are the given morphisms. $L$ becomes a module if we define $r*_L(a,b)=(r*_Xa,r*_Yb)$.

This gives us another element of $L$ by the fact that $R$-module morphisms satisfy $f(rx)=rf(x)$. We may also define the $+$ operation on $L$ as $(x, y)+_L(x', y') = (x+_X x', y+_Yy')$. This gives another element of $L$ by the fact that a morphism of $R$-modules must satisfy $f(x+y) = f(x)+ f(y)$.

Now if we let $X \xleftarrow {\pi_1} L \xrightarrow {\pi_2} Y$ be the projection maps onto the first and second coordinates respectively and we are given morphisms $Y \xleftarrow j A \xrightarrow k X$ satisfying $(g \circ j) (a)= (f \circ k) (a) $ we know that $(j(a),k(a))$ is an element of $L$ by definition. It is also easy to check that the map $$A \xrightarrow p L:a \mapsto (i(a), j(a))$$ is an $R$-module morphism and that the resulting diagram of all of the objects and morphisms mentioned above is commutative. So it follows that $(L, \pi_1, \pi_2)$ is the pullback of the diagram $X \rightarrow M \leftarrow Y$.

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