Differential Geometry: Comparing Metric Tensors

[WWGD]

The proof does not consider the degree of a map $X\to X$, but of a map $X^4\to X^4$. This makes sense because $X^4\cong\mathbb{R}^6$. I think @WWGD had a typo of $X^6$ for $\mathbb{R}^6$ in his post.

Myb bad, thanks for pointing it out , for links and followup. I am editing as we speak.

 

[Infrared]

There is a theorem here that is implicitly assumed which is that the square of any homeomorphism of an orientable manifold is orientation preserving. This is clear for a compact manifold but for a non-compact manifold, how does one prove it?

I think the point is the same as in the closed case. If $X$ is orientable, then all of the local homology groups $H_n(X,X-\text{point})$ are naturally isomorphic (identify the preferred generators), so a homeomorphism $f:X\to X$ with $f(x_1)=x_2$ gives an isomorphism $H_n(X,X-x_1)\to H_n(X,X-x_2)\cong H_n(X,X-x_1),$ and then $f\circ f$ has to induce the identity, so $f\circ f$ preserves orientation.

The only difference from the closed case is that the generators are not induced by a homology class of $X$.

 

[lavinia]

I think the point is the same as in the closed case. If $X$ is orientable, then all of the local homology groups $H_n(X,X-\text{point})$ are naturally isomorphic (identify the preferred generators), so a homeomorphism $f:X\to X$ with $f(x_1)=x_2$ gives an isomorphism $H_n(X,X-x_1)\to H_n(X,X-x_2)\cong H_n(X,X-x_1),$ and then $f\circ f$ has to induce the identity, so $f\circ f$ preserves orientation.

The only difference from the closed case is that the generators are not induced by a homology class of $X$.

Right. Two oriented charts are connected by a path of overlapping oriented charts.