# Show that there exists a pullback square for modules.

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?

• Do you know how pullbacks are defined for sets? – Arnaud D. Jun 13 at 13:12
• Define $L$ as a submodule of $L\oplus Y$ rather than a quotient. – user10354138 Jun 13 at 13:19
• 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? – trisct Jun 13 at 13:22

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 $$X \rightarrow M \leftarrow Y$$.