Proving HomR(F;M) isomorphic to M^n for Free Modules of Rank n < 1

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The discussion centers on proving that HomR(F;M) is isomorphic to M^n for free modules F of rank n < 1, where R is a commutative ring with unity. The proposed mapping, Psi: HomR(F;M) → M^n defined by psi(f) = (f(e1); f(e2); ... ; f(en)), aims to establish this isomorphism. The key challenge identified is demonstrating that this mapping is onto, which can be addressed using the Universal property of free modules.

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View attachment Exampleoflatex.pdf

Let R be a commutative ring with 1. If F is a free module of rank n < 1, then show that
HomR(F;M) is isomorphic to M^n, for each R-module M.

I was thinking about defining a map
Psi : HomR(F;M)--> M^n by psi(f) = (f(e1); f(e2); ... ; f(en))
where F is free on (e1; ... ; en) and
show Psi is an isomorphism. But I am having difficulties showing it is onto.
 
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Oh, I think I can use the Universal property of free modules to get the onto part.
 

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