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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 ?

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  • $\begingroup$ Isn't f being $ D^1 $ enough instead of $ C^1 $? You get the derivative of T from the chain rule. $\endgroup$
    – Zsbán Ambrus
    Commented Mar 24, 2012 at 17:21
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    $\begingroup$ Is there are problem to extend your f on sphere? because here living obvious topological resons, but same time you can approximate f by smooth automorphism of S^n and saying what you already say. $\endgroup$
    – Bad English
    Commented Mar 24, 2012 at 21:33

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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 $\operatorname{deg}(T)= \operatorname{deg}(f\circ f)= \operatorname{deg}(f) \operatorname{deg}(f)$. For an invertible linear transformation, the topological degree is the sign of the determinant, which proves your claim.

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    $\begingroup$ @Pietro Majer: Actually, how do you see that $f$ is proper directly? Instead, one can observe that $f$ is 1-1 and onto and, hence, a homeomorphism by invariance of domain theorem. (Only then I see why $f$ is proper.) Now, of course, the degree argument applies. On the other hand, how do you explain this to an unprepared undergraduate without introducing homology? I can imagine telling him/her about orientation and orientation-reserving/reversing maps and indicating why reversing orientation twice amounts to preserving orientation, etc. $\endgroup$
    – Misha
    Commented Mar 24, 2012 at 15:54
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    $\begingroup$ Yes, $f$ is proper because it's a homeo, and it is a homeo just because $f\circ f $ is so, as explained in David Cohen's answer (now deleted). Topological degree can be introduced by several tools, and it is sometimes taught at undergraduate level. There could be a more elementary proof though. $\endgroup$
    – Pietro Majer
    Commented Mar 24, 2012 at 21:52

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