Prove that the direct sum of any collection of free $R$-modules is free

Question

I've found plenty of answers to this, but none that I can understand, I'm a beginner and struggling. One possible solution involves identifying the union of the bases as a basis, but I cannot quite grasp this concept.

Answer 1

Both the direct sum (coproduct in the category of $R$-modules) and free module conditions are defined by universal properties.

A module $F$ is free on the set $X$ (the basis) if for any other $R$-module $M$ and any set map $\varphi: X \to M$, there is an $R$-module homomorphism $\bar{\varphi}: F \to M$ such that $\varphi = \bar{\varphi} \circ i$, where $i: X \to F$ is the inclusion of the set into the free module $F$. In other words, you say where the free generators go, and you have defined a homomorphism from the free module.

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The direct sum of a collection of modules $\left\{ F_\alpha \right\}_{\alpha \in A}$ is the module $\oplus_{\alpha \in A} F_\alpha$, together with a collection of injections $\iota_{\alpha}: F_{\alpha} \to \oplus_{\alpha \in A} F_\alpha$, such that for any other $R$-module $M$ and $R$-module homomorphisms $\psi_{\alpha}: F_\alpha \to M$, there exists a unique map $\psi: \oplus_{\alpha \in A} F_\alpha \to M$ such that $\psi_{\alpha} = \psi \circ \iota_{\alpha}$. In other words, defining a map from each of a collection of modules to a fixed target module is equivalent to defining a map from their direct sum.

Say that you have a collection $\left\{ F_\alpha \right\}_{\alpha \in A}$ of free $R$-modules. How do we show that the direct sum $\oplus_{\alpha \in A} F_\alpha$ is free? We need to come up with a candidate basis set and check that the universal property holds for this set. Remember that for each $\alpha \in A$, $F_{\alpha}$ is free, so it has a basis set, say $X_{\alpha}$. Let's define the set $$ X = \bigsqcup_{\alpha \in A} X_{\alpha} $$ and show that it is a basis for $\oplus_{\alpha \in A} F_\alpha$. It's clear that we have the set inclusion $i: X \to \oplus_{\alpha \in A} F_\alpha$ by defining $i \big|_{X_{\alpha}} = i_{\alpha}$ for each $\alpha \in A$.

To show that $X$ is a basis, fix an arbitrary $R$-module $M$ and choose a set map $\varphi: X \to M$. This is equivalent to choosing, for each $\alpha \in A$, a map $\varphi_{\alpha}: X_{\alpha} \to M$. Because each $F_{\alpha}$ is free, we have a homomorphism $\bar{\varphi}_{\alpha}: F_{\alpha} \to M$ satisfying $\varphi_{\alpha} = \bar{\varphi}_{\alpha} \circ i_{\alpha}$.

Now, by the universal property of the direct sum, we have a homomorphism $\bar{\varphi}: \oplus_{\alpha \in A} F_\alpha \to M$ satisfying $\bar{\varphi}_{\alpha} = \bar{\varphi} \circ \iota_{\alpha}$. Now, we may conclude that $\varphi = \bar{\varphi} \circ i$, as desired.