Free modules over two sets of the same cardinality are isomorphic

Let R be a ring with 1, and let A and B be sets of the same cardinality. Prove that the free modules F(A) and F(B) are isomorphic.

We begin with a lemma.

Lemma: Let R be a ring with 1 and let \theta : A \rightarrow B be a set map. Note that \iota \circ \theta : A \rightarrow F(B), where F(B) is the free R module on B and \iota : B \rightarrow F(B) is the inclusion map. By the universal prperty of free modules, the induced map \Theta : F(A) \rightarrow F(B) is an R-module homomorphism. If \theta is injective, then so is \Theta, and if \theta is surjective, then so is \Theta. Proof: Suppose \theta is injective, and let x \in \mathsf{ker}\ \Theta. Now \Theta(x) = \Theta(\sum r_i \cdot a_i) = \sum r_i \Theta(a_i) = \sum r_i \theta(a_i) = 0. Since \theta is injective, the \theta(a_i) are distinct. Since F(B) is free on B \supseteq \mathsf{im}\ \theta, we have r_i = 0 for each i. Thus x = 0, and so \mathsf{ker}\ \Theta = 0. So \Theta is injective. Suppose now that \theta is surjective, and let y = \sum r_i b_i \in F(B). Since \theta is surjective, we have b_i = \theta(a_i) for some a_i \in A (not necessarily distinct). Then y = \sum r_i \cdot \theta(a_i) = \Theta(\sum r_i a_i), so that \Theta is surjective. \square

Since in our case \theta is both injective and surjective, \Theta is bijective, hence an R-module isomorphism.

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