Recently a student asked me the following (elementary looking) question :


If $T$ is an invertible linear transformation of some finite-dimensional space $E$ into itself which factorizes as $T = f \circ f $ where $f : E \mapsto E$ is *continuous*, must $T$ have positive determinant ?


Of course this is trivially true if $f$ is itself linear. It is also an easy exercise to show that this also holds when $f$ behaves locally like a linear transformation, that is, when it is $C^1$ : $T$ then factorizes as $T = df\_{f(0)} \circ df\_0 $, and since $x \mapsto \det df\_x$ keeps a constant sign, we're done.


When $f$ is only continuous, this certainly still holds but I suspect this requires rather deep properties of continuous maps (unless I missed something obvious ...) with which I'm not very familiar. Hence two questions :


1) Is there an "elementary" proof of this ? (in which case I apologize for this question)


2) Does this property sound obvious to experts ? That is, is there some two-lines proof of this with a sufficient background ? If yes, what would be good references (books for example) to acquire this background ?



Square of a continuous map
The first relevant fact about $f$ is that it is a [proper map](http://en.wikipedia.org/wiki/Proper_map). 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.


