Square of a continuous map

2012-03-24T05:39:37Z

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 ?

Answer by Pietro Majer for Square of a continuous map

2012-03-24T06:38:28Z

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.

Square of a continuous map - MathOverflow

Square of a continuous map

Answer by Pietro Majer for Square of a continuous map