MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
Isn't f being $ D^1 $ enough instead of $ C^1 $? You get the derivative of T from the chain rule. – Zsbán Ambrus Mar 24 '12 at 17:21
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. – Bad English Mar 24 '12 at 21:33
up vote 23 down vote accepted

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.

share|cite|improve this answer
@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. – Misha Mar 24 '12 at 15:54
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. – Pietro Majer Mar 24 '12 at 21:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.