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