Lemma 10.87.4. Let $f : M \to N$ and $g : M \to M'$ be maps of $R$-modules. Consider the pushout of $f$ and $g$,

Then $g$ dominates $f$ if and only if $f'$ is universally injective.

Lemma 10.87.4. Let $f : M \to N$ and $g : M \to M'$ be maps of $R$-modules. Consider the pushout of $f$ and $g$,

\[ \xymatrix{ M \ar[r]_ f \ar[d]_ g & N \ar[d]^{g'} \\ M' \ar[r]^{f'} & N' } \]

Then $g$ dominates $f$ if and only if $f'$ is universally injective.

**Proof.**
Recall that $N'$ is $M' \oplus N$ modulo the submodule consisting of elements $(g(x), -f(x))$ for $x \in M$. From the construction of $N'$ we have a short exact sequence

\[ 0 \to \mathop{\mathrm{Ker}}(f) \cap \mathop{\mathrm{Ker}}(g) \to \mathop{\mathrm{Ker}}(f) \to \mathop{\mathrm{Ker}}(f') \to 0. \]

Since tensoring commutes with taking pushouts, we have such a short exact sequence

\[ 0 \to \mathop{\mathrm{Ker}}(f \otimes \text{id}_ Q ) \cap \mathop{\mathrm{Ker}}(g \otimes \text{id}_ Q) \to \mathop{\mathrm{Ker}}(f \otimes \text{id}_ Q) \to \mathop{\mathrm{Ker}}(f' \otimes \text{id}_ Q) \to 0 \]

for every $R$-module $Q$. So $f'$ is universally injective if and only if $\mathop{\mathrm{Ker}}(f \otimes \text{id}_ Q ) \subset \mathop{\mathrm{Ker}}(g \otimes \text{id}_ Q)$ for every $Q$, if and only if $g$ dominates $f$. $\square$

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