The first relevant fact about $f$ is that it is a [proper map][1]. In such a situation the topological (Brouwer) degree of $f$ is well-defined, and by the product rule $\operatorname{deg}(T)= \operatorname{deg}(f\circ f)= \operatorname{deg}(f) \operatorname{deg}(f)$. For an invertible linear transformation, the topological degree is the sign of the determinant, which proves your claim.

[1]:http://en.wikipedia.org/wiki/Proper_map