Proof $F$ is a free module on $X$

Question

Let $\{X_i:i\in I\}$ be a collection of mutually disjoint sets and for each $i \in I$ let $F_i$ be a free module on $X_i$ with $l_i: X_i\rightarrow F_i$. Let $X=\bigcup_{i\in I}X_i$ and $F=\sum_{i\in I}F_i$ with $\phi_i:F_i\rightarrow F$ canonical injection. Define $l:X\rightarrow F$ by $l(x)=\phi_i \circ l_i(x)$ for $x\in X_i$. Prove that $F$ is a free module on $X$.

Answer 1

We assume the $F_i$ are free over a common ring R. Clearly $X$ is a generating set for $F$, we must show it is linearly independent. Suppose that $$\sum_{\substack{ij\\finite}}a_{ij}x_{ij}=0$$ for $a_{ij}\in R$ and $x_{ij}\in X_i$ distinct. Since $X_i$ are all disjoint it follows that $F_i\not\in\text{Span}(F_j)_{j\in I\backslash \{i\}},$ so $$\sum_{i\in I} F_i=\bigoplus_{i\in I}F_i.$$ It follows that for all $i$ we must have $$\sum_{j}a_{ij}x_{ij}=0,$$ but then since $F_i$ is free with basis $X_i$ it follows that $a_{ij}=0$ for all $i,j.$ Then $X$ is a basis for $F$ over $R,$ hence $F$ is free.

Answer 2

Here's an abstract nonsense approach. As the sets $X_i$ are mutually disjoint, the union $X = \bigcup X_i$ is the coproduct of the $X_i$ in the category of sets. Essentially by definition of a free object, the free functor $Free: Sets \longrightarrow R-Mod$ is left adjoint to the forgetful functor $Forget: R-Mod \longrightarrow Sets$. Indeed, the universal property immediately gives you such a natural bijection. It is known that left adjoint functors are cocontinuous (preserve colimits). Thus, $Free(X) = \bigoplus Free(X_i)$, and the result follows.