Let $M_0$ be an R-module, and suppose $M_{n+1}$ is the pushout of the diagram below as shown, for all $n \in \mathbb{N}$:

$$\begin{array}{ccc}M_n&\to& M_{n+1}\\\uparrow &&\uparrow\\A&\to& B\end{array}$$

Define $M:= colim(M_0 \rightarrow M_1 \rightarrow M_2 \rightarrow.....)$

Can we deduce that $M_0$ is a submodule of $M$? What about if the left arrow is surjective and the bottom arrow is injective? Would this help?

I am asking this in order to complete my understanding of the proof that $R-mod$ has enough injectives. In this proof, A and B are given, however there is no justification given for the claim which I am inquiring about, it is just stated to be obvious. Thanks.


If not yet done, convince yourself (by element chasing) that the pushout of an injective module homomorphism is injective.

Then, if $A\to B$ is injective, then $M_i$ embeds to $M_{i+1}$, and it's easy to see that $M':=\bigcup_iM_i$ satisfies the universal property of the colimit, hence $M\cong M'$ and the natural arrow $M_i\to M$ is injective for all $i$.

  • $\begingroup$ Thanks, this is exactly what I was looking for! $\endgroup$ – Daven Jan 3 at 15:58

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