Something I'm reading says that for a commutative ring $A$ and $f\in A$, a free resolution of $A_f$ is $$ 0 \to \bigoplus\nolimits_{n \in \mathbf{N}} A \to \bigoplus\nolimits_{n \in \mathbf{N}} A \to A_f \to 0 $$ where the first map sends the $(x_0, x_1, \ldots)$ to $(fx_0 - x_1, fx_1 - x_2, \ldots)$ and the second map sends $(x_0, x_1, \ldots)$ to $x_0 + x_1/f + x_2/f^2 + \ldots$.
My question is that the composite of the two maps seems to be $fx_0$, so this it is not a free resolution, right? I probably am making a dumb mistake.
The first map should be: $$(x_0, x_1, x_2, \dots) \mapsto (x_0, x_1 - fx_0, x_2 - f x_1, x_3 - f x_2 \dots)$$ Then the composite becomes: $$(x_0, x_1, x_2, \dots) \mapsto x_0 + (x_1 - fx_0)/f + (x_2 - fx_1)/f^2 + \dots =0.$$ The first map is clearly injective, and the second map is clearly surjective. It's not hard to check that the second map has a kernel equal to the image of the first map.