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Given module homomorphism $\alpha:M\to N$ and $P\le M$ (submodule). Can we say $\alpha(P)\le \alpha(M)$? In the solution the lecturer points out that the image of $\alpha$ is the same as the image of $\alpha\circ i$ where $i$ is the inclusion map $i:P\to M$. How does this help us see $\alpha(P)\le \alpha(M)$?

If the inclusion map is not relevant to showing $\alpha(P)\le \alpha(M)$, perhaps it's needed in some way to show $\alpha(P)\leq N $? (please see comment below)

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@Joe Johnson. Ah this result is very obvious indeed we already know that Im($\alpha$) is a submodule of N, thus a module in its own right, also P is a subset of M, this is what is required for $\alpha (P)\le\alpha(M)$. I'm trying to identify what is the role of the inclusion map in my lecturer's solution? Maybe it's relevant to the rest of the question which asks to show $\alpha(P)\le N$? It seems to me redundant because we know $\alpha(M)\le N$, so with the knowledge $\alpha (P)\le\alpha(M)$ from what you said we can conclude $\alpha (P)\le N$ – Catherine Nov 24 '11 at 14:33
Your lecturer is saying that $\alpha$ restricted to $P$ has the same image as $\alpha\circ i$. Since $i$ and $\alpha$ are both homomorphisms, $\alpha(P)$ is the image of a homomorphism and is therefore a submodule of $N$. – Joe Johnson 126 Nov 24 '11 at 15:10
@JoeJohnson126, thank you this finally makes sense! That inclusion map was so puzzling to me. – Catherine Nov 24 '11 at 15:29
up vote 2 down vote accepted

First, you need to show that the image of a module under a homomorphism is also a module. Once you have shown that then $\alpha(P)$ is certainly a subset of $\alpha(M)$. Also, $P$ and $M$ are both modules. Also, $\alpha(P)$ and $\alpha(M)$ are both modules by the first step.

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