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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.

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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 $i$.

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

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