Proof: Suppose [math]f[/math] and [math]g[/math] are isometries which means they are onto, one-to-one, and distance-preserving. Define the composition of [math]f[/math] and [math]g[/math] as [math]f\left(g\left(x\right)\right)[/math].[br][br]A function is said to be onto if there is a point, x, in the domain for which [math]f\left(x\right)=y[/math] where any [math]y[/math] is an element of the codomain. In the relation to the composition, [math]f\left(g\left(x\right)\right)[/math], there needs to be a point [math]x[/math] in the domain for which [math]f\left(g\left(x\right)\right)=y[/math] where [math]y[/math] is in the codomain. Since we know that [math]g\left(x\right)[/math] is an isometry, we know that [math]g\left(x\right)[/math] is onto and any point [math]x[/math] maps to a point [math]y[/math]. We also know that [math]f[/math] is onto, so any point in the domain maps to the codomain. Therefore, the composition of [math]f[/math] and [math]g[/math] is onto.[br][br]A function is said to be one-to-one if [math]x_1=x_2[/math] implies [math]f\left(x_1\right)=f\left(x_2\right)[/math]. Pick two points [math]x_1[/math] and [math]x_2[/math]. If [math]f\left(g\left(x_1\right)\right)=f\left(g\left(x_2\right)\right)[/math], then it is implied that [math]g\left(x_1\right)=g\left(x_2\right)[/math] because [math]f[/math] is one-to-one by definition. Notice if [math]g\left(x_1\right)=g\left(x_2\right)[/math], [math]x_1=x_2[/math] because [math]g[/math] is one-to-one by definition. Since [math]x_1=x_2[/math], we can conclude that the composition [math]f\left(g\left(x\right)\right)[/math] is one-to-one.[br][br]A function is said to be distance-preserving if the distance between to points [math]A[/math] and [math]B[/math] is the same as the distance between the images, [math]f\left(A\right)[/math] and [math]f\left(B\right)[/math]. Pick points [math]X[/math] and [math]Y[/math]. Notice, [math]\left|XY\right|=\left|g\left(X\right)g\left(Y\right)\right|[/math] because [math]g[/math] is distance preserving by definition. Also notice, [math]\left|g\left(X\right)g\left(Y\right)\right|=\left|f\left(g\left(X\right)\right)f\left(g\left(Y\right)\right)\right|[/math] because [math]f[/math] is also distance preserving by definition. Since we know these two things, we can conclude that [math]\left|XY\right|=\left|f\left(g\left(X\right)\right)f\left(g\left(Y\right)\right)\right|[/math]. Therefore, the composition of [math]f[/math] and [math]g[/math] is distance-preserving.[br][br]Since [math]f\circ g[/math] is onto, one-to-one, and distance-preserving, we can conclude that the composition of [math]f[/math] and [math]g[/math] is an isometry if [math]f[/math] and [math]g[/math] are isometries. [math]\diamondsuit[/math]