Let $R$ be a ring with identity, $I$ ($\neq R$) an ideal of $R$, $F$ a free $R$-module with basis $X$ and $\pi : F\to F/IF$ the canonical epimorphism. Then $F/IF$ is a free $R/I$ module with basis $\pi (X)$ and $|\pi (X)| = |X|$.
Recall that $IF = \{\sum_{i=1}^n r_ia_i \mid r_i\in I, a_i \in F, n \in \Bbb{N}\}$ and that the action of $R/I$ on $F/ IF$ is given by $(r+I)(a+IF) = ra +IF$.
Proof: ……… Finally if $x, x' \in X$ and $\pi (x) = \pi (x')$ in $F/IF$, then $(1_R+I)\pi (x)-(1_R+I)\pi (x’)=0$. If $x\neq x'$, the preceding argument implies that $1_R\in I$, which contradicts the fact that $I\neq R$. Therefore, $x = x'$ and the map $\pi: X \to \pi (X)$ is a bijection, whence $|X|= |\pi(x)|$·
Question: I don’t get Author proof of $\pi|_X$ is injective. It seems like Author beforehand assuming injectivity. For context, preceding argument refer to $\pi (X)$ is linearly independent.
Following argument is not circular to prove $\pi|_X$ is injective: if $x\neq x’$ and $\pi(x)=\pi (x’) =x+IF=x’+IF$, then $x-x’\in IF$. So $x-x’=\sum_{i=1}^nr_ia_i$ where $r_i\in I$, $a_i\in F$ and $n\in \Bbb{N}$. Since each $a_i$ is linear combination of elements of $X$ and $I$ is an ideal, we have $$x-x’=\sum_{i=1}^nr_ia_i =\sum_{i=1}^m s_i b_i$$ where $s_i\in I$, $b_i\in X$ and $m\in \Bbb{N}$. By linear independence of $X$, $m=2$, $(s_1,b_1)=(1_R,x)$ and $(s_2,b_2)=(-1_R,x)$. That means $s_1=1_R\in I$. Thus we reach contradiction. Hence $x=x’$.