Write down a partition $X^*$ of the unit square $X=[0,1]\times[0,1]$ such that the quotient space is the Klein bottle.
I understand the definition of Quotient topology and Partitions, however, don't fully understand how to apply this.
Write down a partition $X^*$ of the unit square $X=[0,1]\times[0,1]$ such that the quotient space is the Klein bottle.
I understand the definition of Quotient topology and Partitions, however, don't fully understand how to apply this.
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Perhaps you know the traditional way to construct a Klein bottle from the square $[0,1] \times [0,1]$:
Gluing Rule 1: Glue the bottom edge $[0,1] \times 0$ to the top edge $[0,1] \times 1$ without any twist, thereby gluing each point $(x,0)$ to the point straight above it $(x,1)$.
Gluing Rule 2: Glue the left edge $0 \times [0,1]$ to the right edge $1 \times [0,1]$ with a twist, thereby gluing each point $(0,y)$ to the point $(1,1-y)$.
These rules for gluing define a relation $\sim$ from the square $[0,1] \times [0,1]$ to itself: $(x,0) \sim (x,1)$ for each $x \in [0,1]$, and $(0,y) \sim (1,y-1)$ for each $y \in [0,1]$.
So given a point $(x,y) \in [0,1]$, to write the partition element containing $(x,y)$ you ask: What points is $(x,y)$ related to? And what points are those related to? And what points are those related to? And you keep going until you don't get any more points. The collection of points that you get by this process is the equivalence class that contains $(x,y)$. So let's work in cases.
Case 1: Given the point $(0,0)$. It is related to both $(0,1)$ and $(1,1)$. These are both related to $(1,0)$. And now there are no new points. So you get one equivalence class $$\{(0,0),(0,1),(1,0),(1,1)\} $$
Case 2: Given the point $(x,0)$ with $0<x<1$. It is related to $(x,1)$. And now there are no new points. So you get one equivalence class $$\{(x,0),(x,1)\} $$ for each $0<x<1$.
Case 3: Given the point $(0,y)$ with $0<y<1$. It is related to $(1,1-y)$. And now there are no new points. So you get one equivalence class $$\{(0,y),(1,1-y)\} $$ for each $0<y<1$.
Case 4: Given the point $(x,y)$ with $0<x<1$ and $0<y<1$. It is not related to anything. So you get one equivalence class $$\{(x,y)\} $$ for each $0<x<1$, $0<y<1$.