I am trying to solve the following exercise from the book Differential Geometry by Loring W. Tu:
5.4 Total curvature
The total curvature of a smooth oriented surface $M$ in $\mathbb{R}^3$
is defined to be the integral $\int_M K$ if it exists, of the Gaussian curvature $K$. Prove that the total curvature of $M$ is, up to sign, the area of the image of the Gauss map:
$$\int_M K = \text{Area of} \ \nu(M).$$
I'm pretty sure that surfaces are assumed to be connected, otherwise this theorem is clearly false. (Consider two disjoint spheres in $\mathbb{R}^3$ with outward unit normal.) Even then I am not sure that this is true as stated. I know how to prove it in the case when $\nu$ is injective on the set of points where $K$ is non-zero. In this case the restriction of $\nu$ to this set is actually a diffeomorphism onto its image and if $dV_g$ and $dV$ are the volume forms of $M$ and $\mathbb{S}^2$ respectively, then the pullback of $dV$ with $\nu$ is just $KdV_g$. My idea to solve the general case was to introduce a partition of unity with respect to an open cover such that on each element of the cover $\nu$ is a diffeomorphism onto its image, but I run into trouble with this approach because of the possible non-injectivity of $\nu$. In fact when I brought this up with one of my professors, we also didn't know how to proceed and in turn tried to consider some examples. This discussion culminated in what we think is a counterexample which I will now describe.
Consider the following subsets of $\mathbb{R}^3$: $A=${$(x,y,z)\in \mathbb{R}^3 \vert x^2+y^2 \geq 2\, z=0$} and $B=${$(x,y,z)\in \mathbb{R}^3 \vert x^2+y^2+(z-1)^2 = 1, z \geq 1$}. So $A$ is the plane $z=0$ with the open disk of radius $2$ removed and $B$ is the upper hemisphere of a unit sphere around the point $(0,0,1)$. We can then imagine connecting these two in such a way that we get a smooth surface embedded in $\mathbb{R}^3$. The image of the Gauss map of this surface clearly has positive volume no matter what unit normal vector field we pick, since it either contains the whole upper hemisphere or the whole lower hemisphere of $\mathbb{S}^2$. Now consider the circle with radius $2$ in the plane $z=0$ parameterized anti-clockwise with unit speed such that it winds around only once and denote this parameterization with $\gamma$. If we denote with $\Omega$ the surface it encloses and denote with $\kappa_N$ it's curvature with respect to the inward pointing normal relative to the plane $z=0$ then the Gauss-Bonnet formula (John M. Lee - Introduction to Riemannian Manifolds theorem 9.3) yields:
$$\int_{\Omega}KdA + \int_{\gamma} \kappa_{N} ds = 2\pi.$$
And since the second integral is equal to $2\pi$ we get $\int_{\Omega}KdA=0$, so this integral can't be equal (up to sign) to the area of the image of the Gauss map which is positive. This result is also independent on how we connected $A$ and $B$ to a smooth connected surface in $\mathbb{R}^3$. I think the problem is that no matter how we connect $A$ and $B$ we won't be able to do it in a way that would make the Gauss map injective.
The only answer on here that asks a similar question is this, but as I said, I understand this case. I want to know if the counterexample works and if it does are there even easier counterxamples? If the claim is actually true as stated, then I would like to see an argument why it is true.
EDIT:
It has been confirmed in the comments that this exercise is wrong as written and I have also located an errata for Tu's book on his website which adds that the Gauss map should be injective on points where $K\neq 0$ and that $K$ should be either non-negative or non-positive everywhere. This is so that the Gauss map is either orientation-preserving or orientation-reversing on the set where $K\neq 0$. @TedShifrin also provided an easier counterexample: the torus in $\mathbb{R}^3$.