Square of a continuous map - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:52:46Z http://mathoverflow.net/feeds/question/92061 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92061/square-of-a-continuous-map Square of a continuous map js 2012-03-24T05:39:37Z 2012-03-24T08:47:26Z <p>Recently a student asked me the following (elementary looking) question :</p> <p>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 <em>continuous</em>, must \$T\$ have positive determinant ?</p> <p>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.</p> <p>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 :</p> <p>1) Is there an "elementary" proof of this ? (in which case I apologize for this question)</p> <p>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 ?</p> http://mathoverflow.net/questions/92061/square-of-a-continuous-map/92062#92062 Answer by Pietro Majer for Square of a continuous map Pietro Majer 2012-03-24T06:38:28Z 2012-03-24T08:47:26Z <p>The first relevant fact about \$f\$ is that it is a <a href="http://en.wikipedia.org/wiki/Proper_map" rel="nofollow">proper map</a>. 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.</p>