Simplicial complexes

(1) it has the following triangulation. The vertices with the same label are identified.
We check that the 2-simplices {1,2,4},{1,2,7},{1,2,8},{1,3,4},{1,3,5},{1,3,6},{1,5,6},{1,7,8},{2,3,5},{2,3,7},{2,3,8},{2,4,5},{3,4,8},{3,6,7},{4,5,6},{4,6,8},{6,7,8} are distinct.
The following two are not triangulation
(2) We prove for any finite simplicial complex $K$, $|K|$ is locally connected[any point x ∈ |K| and any open set U containing x, there is a connected open set V such that x ∈ V ⊆ U].
Let $x∈{|K|}$ and suppose that $U$ is an open set containing $x$. Let $D$ be the disjoint union of simplices from which $|K|$ is obtained, and let $q:D→{|K|}$ be the quotient map. The restriction of $q$ to the insides of the simplices in $D$ is a bijection, so there is exactly one simplex $σ$ in $D$ such that $x∈q(\operatorname{inside}(σ))$.
$q^{-1}(x)∈q^{-1}(U)∩\operatorname{inside}(σ)$ is open in $σ⇒q(q^{-1}(U)∩\operatorname{inside}(σ))∋x$ is connected, open in $|K|$.

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We prove the comb space $X$ is not locally connected. (0,1) ∈ X. Suppose the open set $B((0,1),1)$ contains an open set $U$, let $p_xU$ be the x-projection of $U$. Let $n$ such that $\frac1n∈p_xU$, then $p_xU$ is disconnected by $\Set{x∈p_xU:x≤\frac1{n+1}}$ and $\Set{x∈p_xU:x≥\frac1n}$, so $U$ is disconnected.
Note. $X$ is path connected but not locally connected. After $\{0\}×[0,1]$ removed, the space$$\{(x, y): 0≤y≤1, \text { and } x=1/n, \text { for some } n∈ℕ\}∪([0,1]×\{0\})$$is locally connected[as $p_xB\left((\frac1n,1),\frac1n-\frac1{n+1}\right)=\Set{\frac1n}$] and path connected. The $N$-simplex spanned by $a_0,…,a_n$ is the set of all $x∈ℝ^N$ such that $x=\sum_{i=0}^n λ_ia_i$, where $\sum_{i=0}^n λ_i=1,λ_i≥0∀i$. The numbers $λ_i$ are uniquely determined by $x$.
Define a function $t_{a_i}:{|K|}→[0,1]$ by $t_{a_i}(x)=λ_i$ if $x∈τ$ and 0 if $x∉τ$. Then $t_{a_i}$ is continuous.
$|K|$ is Hausdorff: Let $x,y∈{|K|},x≠y$, so ∃ vertex $v:t_v(x)≠t_v(y)$. Take $r∈ℝ$ between $t_v(x),t_v(y)$. Then $t_v^{-1}[0,r),t_v^{-1}(r,1]$ are disjoint open neighborhood of $x,y$.

Surfaces

$ℝ^2∖\{0\}$ is not closed, by sheet3 Q5, $Y/\mmlToken{mi}∼$ is not Hausdorff.

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The map $f:X_1→X_2,f((x,y)_1)=(x,y)_2$ is a homeomorphism.
For any $x∈X_1$ [including $(0,0)_1$], let $U∋x$ open in $X_1$ [$U$ possibly contains $(0,0)_1$], then $U≅q(U)∋q(x)$ and $q^{-1}(q(U))=U∪f(U∖\{(0,0)_1\})$ is open in $Y$, so $q(U)$ is open in $Y/\mmlToken{mi}∼$.
Similarly, for any $x∈X_2$, $q(x)$ is contained in an open set $q(U)$ homeomorphic to $U$ an open set of $X_2$. Take a triangulation of the sphere. Then glue two of the vertices together in such a way that the result is still a simplicial complex. This can't be done for the tetrahedron since every pair of vertices share an edge, but is possible for larger triangulations.
Any sufficiently small neighborhood can be disconnected by removing the point, whereas in a disk no point can be removed to make it disconnected. So the result is not a closed combinatorial surface.
It clearly satisfies (1)(3). Every 1-simplex is a 1-simplex of the original triangulation, so is a face of precisely two 2-simplices, so (2) is satisfied.

1. Let $C_1$ be the horizontal line $\{\{(y,0),(1-y,1)\}:y∈(0,1)\}$. Then $K∖C_1=$
2. Let $C_2$ be the vertical line $\{\{(0,y),(1,y)\}:y∈(0,1)\}$. Then $K∖C_2=$
3. Let $C_3$ be the dashed line
   $K∖C_3=$ We have to glue the red arrows

[regular hexagon can tessellate $ℝ^2$ by translation in 2 directions, so this torus is a quotient of $ℝ^2$ by translation in 2 directions]
It's homeomorphic to a torus, we can see it directly

Alternatively, label the edges by $x,y,z$, represent the surface by $yzx^{-1}y^{-1}z^{-1}x$ and use Lemma 5.23 \[y{\color{red}\underbrace{zx^{-1}}_\text{swap}}y^{-1}z^{-1}x=yx^{-1}zy^{-1}z^{-1}x=yx^{-1}zy^{-1}(x^{-1}z)^{-1}\text{ is a torus}\] We assumed $𝕊^2∖C$ have two components so it is a disjoint union $𝕊^2∖C=E⊔E'$ where $E,E'$ are connected.

1. Since the interior of every 2-simplex $σ_2$ is connected, either $\mathring{σ_2}⊂E$ or $\mathring{σ_2}⊂E'$. If $\mathring{σ_2}⊂E$ then $σ_2⊂\bar E$ [since $σ_2$ is closed].
2. $σ_2$ is the only 2-simplex in $\bar{E}$. If $\bar E$ contain other vertices, then it lies in $C$ (otherwise $\bar E$ would have other 2-simplices) so $σ_2=\bar{E}$.
3. Let $v_1,v_2$ be vertices of $σ_1$. Replacing $σ_1$ with two 1-simplices $v_1,v$ and $v,v_2$, we get a simplicial circle $C'$, and $\bar{F}$ contains fewer 2-simplices (all 2-simplices of $\bar E$ except $σ_2$), apply induction hypothesis to $\bar F$: There is a homeomorphism from $\bar{F}$ to $𝔻^2$, which takes $C'$ to the boundary circle $∂𝔻^2$.

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   $\bar E=\bar F∪σ_2$. Since gluing two disks along an edge gives a disk, $\bar E$ is homeomorphic to $𝔻^2$.
4. WLOG one of edges $(v,v_1)$ of $σ_2$ is not on $C$ (if both lie on $C$ ⇒ part 2)
   Since $C$ is a simplicial circle, we can list its 1-simplices in order $σ^1_1,…,σ^m_1$ where $σ_1=\{v_1,v_2\}$ and $∃i$ such that $σ^i_1$ ends at $v$. We get two simplicial circles $C_1=\{v,v_1\}∪σ^1_1∪⋯∪σ^i_1$ and $C_2=σ^{i+1}_1∪⋯∪σ^m_1∪\{v_1,v\}$. They have the following properties:
   • $𝕊^2∖C_k$ has two components $F_k$
   • one of $F_k$ is a subset of $E$
   • $\bar{F}_k$ has fewer 2-simplices than $\bar{E}$

By induction hypothesis, $\bar{F}_k\,(k=1,2)$ are homeomorphic to $𝔻^2$ and we have $\bar{F}_1∪\bar{F}_2=\bar{E}$.
$\bar{F}_1∩\bar{F}_2=\{v,v_1\}$, so $\bar{F}_1∪\bar{F}_2≅𝔻^2$, since gluing two disks along an edge is homeomorphic to a disk.