Define the connect sum $R \# S$. Prove that $χ(R \# S)=χ(R)+χ(S)-2$. [You may assume that surfaces admit triangulations.]

If $V$ is the number of 0-cells, $E$ the number of 1-cells, $F$ the number of 2-cells, then the Euler characterisic is $χ=V-E+F$. If $S$ is orientable, it is a sphere with a finite number $g ≥ 0$ of handles attached. It has $χ=2-2 g$. If $S$ is non-orientable, it is a sphere with a finite number $h ≥ 1$ of Möbius bands attached. It has $χ=2-h$.

The connect sum is obtained by removing a disc from each surface and then identifying the two circular boundaries (this determines the surface up to homeomorphism).

Pick triangulations for $S_1, S_2$. Remove a triangle from each. Identify the boundaries of the two triangles by identifying three edges and three vertices (up to homeomorphism, it does not matter how we choose to identify these). New number of vertices: $V=V_1+V_2-3$, edges; $E=E_1+E_2-3$, faces: $F=F_1+F_2-2$. So $V-E+F=χ(S_1)+χ(S_2)-2$

The connect sum is obtained by removing a disc from each surface and then identifying the two circular boundaries (this determines the surface up to homeomorphism).

Pick triangulations for $S_1, S_2$. Remove a triangle from each. Identify the boundaries of the two triangles by identifying three edges and three vertices (up to homeomorphism, it does not matter how we choose to identify these). New number of vertices: $V=V_1+V_2-3$, edges; $E=E_1+E_2-3$, faces: $F=F_1+F_2-2$. So $V-E+F=χ(S_1)+χ(S_2)-2$

Describe a cellular decomposition for $H$, and then determine the surfaces $H$ and $H \# H$ in the classification of part (a).

The filled polygon is a 2-cell. The three pairs of identified edges are 1-cells. These bound three pairs of identified vertices which are 0-cells. So $χ(H)=3-3+1=1$. So $H$ is a non-orientable surface obtained from a sphere by attaching a Möbius band (namely $ℝ P^2$).

$H \# H$ has $χ=1+1-2=0$. By the classification, it is either a torus $(g=1)$ or a Klein bottle $(h=2)$. Now $H$ contains a Möbius band $M$ since $χ$ is odd (alternatively, consider the rectangle joining two diametrically opposite sides). We can pick the discs that we remove in the construction of $H \# H$ to avoid this $M ⊂ H$, so $H \# H$ also contains a copy of $M$ so it is non-orientable. So $H \# H$ is a Klein bottle.

$H \# H$ has $χ=1+1-2=0$. By the classification, it is either a torus $(g=1)$ or a Klein bottle $(h=2)$. Now $H$ contains a Möbius band $M$ since $χ$ is odd (alternatively, consider the rectangle joining two diametrically opposite sides). We can pick the discs that we remove in the construction of $H \# H$ to avoid this $M ⊂ H$, so $H \# H$ also contains a copy of $M$ so it is non-orientable. So $H \# H$ is a Klein bottle.

If there were infinitely many points in $φ^{-1}(s)$, then by compactness of $R$ we could find an accumulation point $r_n → r$ in $R$, with $φ(r_n)=s$. By continuity also $φ(r)=s$. This contradicts that there is an open neighbourhood $V_i$ containing $r$ and bijective (indeed diffeomorphic) via $φ$ with a neighbourhood $V$ of $s$.

The number of points $d$ is locally constant (hence constant since $S$ is connected) because $φ: V_1 ⊔ ⋯ ⊔ V_d → V$ is bijective when restricted to each copy $V_i$.

The number of points $d$ is locally constant (hence constant since $S$ is connected) because $φ: V_1 ⊔ ⋯ ⊔ V_d → V$ is bijective when restricted to each copy $V_i$.

Theorema Egregium: The Gaussian curvature $K$ only depends on the Riemannian metric.

Gauss-Bonnet: for a compact surface $S$ (without boundary), $χ(S)=\frac{1}{2 π} ∫_S K d A$.

The metric $J$ on $R$ is $T_r R × T_r R → ℝ, J(v, w)=I(D_r φ ⋅ v, D_r φ ⋅ w)$ (which is an inner product since $I$ is and $D_r φ$ is a linear bijection).

$φ$ is an isometry (i.e. preserves lengths of curves) by the chain rule: if $γ$ is a curve in $R$, then its length is $L(γ)=∫ \sqrt{ J(γ', γ')} d t=∫ \sqrt{ I(D φ ⋅ γ', D φ ⋅ γ')} d t=∫ \sqrt{I(ν', ν')} d t=L(ν)$ where $ν=φ ∘ γ$ is the image curve in $S$.

Since locally isometric surfaces have locally equal $K$, the Gaussian curvatures on $V_i$ and $V$ are the same. So $∫_{V_1 ⊔ ⋯ ⊔ V_d} K d A=d ⋅ ∫_V K d A$. It follows that globally $∫_R K d A=$ $d ⋅ ∫_S K d A$, and hence $χ(R)=d ⋅ χ(S)$ by Gauss-Bonnet.

Gauss-Bonnet: for a compact surface $S$ (without boundary), $χ(S)=\frac{1}{2 π} ∫_S K d A$.

The metric $J$ on $R$ is $T_r R × T_r R → ℝ, J(v, w)=I(D_r φ ⋅ v, D_r φ ⋅ w)$ (which is an inner product since $I$ is and $D_r φ$ is a linear bijection).

$φ$ is an isometry (i.e. preserves lengths of curves) by the chain rule: if $γ$ is a curve in $R$, then its length is $L(γ)=∫ \sqrt{ J(γ', γ')} d t=∫ \sqrt{ I(D φ ⋅ γ', D φ ⋅ γ')} d t=∫ \sqrt{I(ν', ν')} d t=L(ν)$ where $ν=φ ∘ γ$ is the image curve in $S$.

Since locally isometric surfaces have locally equal $K$, the Gaussian curvatures on $V_i$ and $V$ are the same. So $∫_{V_1 ⊔ ⋯ ⊔ V_d} K d A=d ⋅ ∫_V K d A$. It follows that globally $∫_R K d A=$ $d ⋅ ∫_S K d A$, and hence $χ(R)=d ⋅ χ(S)$ by Gauss-Bonnet.