Lemma 10.88.4. 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 10.88.4. 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 $f$. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)