3
$\begingroup$

In Lee's book Introduction to Topological Manifolds, he discusses elementary operations on polygonal presentations. Before the question, here are the terminologies that I am going to be using:


  • A polygonal presentation, written $\mathcal{P}=\left<S\mid W_1,\dots ,W_k\right>$, is a finite set $S$ together with finitely many words $W_1,\dots ,W_k,$ in $S$ of length $3$ or more, such that every symbol in $S$ appears at least once in one word. (He also discusses the special case where $S$ is a singleton set and there is only one word of length $2$, but since it is irrelevant to this question, I will omit it.)

  • The geometric realization $\mid \mathcal{P} \mid$ of a given polygonal presentation $\mathcal{P}=\left<S\mid W_1,\dots ,W_k\right>$ is defined as follows: For each $W_i$, let $P_i$ denote the regular $n_i$-gon in the plane.(Here $n_i$ is the length of $W_i$.) Define a one-to-one correspondence between the symbols of $W_i$ and the edges of $P_i$ in counterclockwise order, and then let $\sim$ be the smallest equivalence relation $P=\coprod P_i$ that identifies the edges that have the same label. We define $\mid \mathcal{P}\mid$ to be the quotient space $P/\sim$.

  • Two polygonal presentations are said to be equivalent if their geometric realizations are homeomorphic.


Let us get back to the question. In the book, the author claims that certain operations on polygonal presentations yield equivalent presentations. For instance, the operation called folding, which is
$$\mathcal {P}=\left<S\mid W_1ee^{-1},\dots ,W_k\right> \mapsto \mathcal{P}'=\left<S,e\mid W_1,\dots ,W_k\right>,$$ produces equivalent presentations. The basic idea of the proof of why this is true is outlined in the book; it goes like this:

Let $q:P=\coprod P_i \to \mid\mathcal{P} \mid$ and $q':P'=\coprod P'_i\to \mid\mathcal{P}' \mid$ are the quotient maps corresponding to $\mathcal {P}$ and $\mathcal {P}'$. Choose a quotient map $f:P\to P'$that identifies only the edges labeled as $e$. Then $q$ and $q'\circ f$ are quotient maps that makes the same identifications, so their images $\mid\mathcal{P} \mid$ and $\mid\mathcal{P}' \mid$ are homeomorphic to each other.

I can understand the basic idea, but I am not sure why $q$ and $q'\circ f$ make the same identifications, especially when it is quite difficult to explicitly write out the equivalence relation that determines $q$ and $q'$. Can anyone tell me how to rigorously show that $q$ and $q'\circ f$ make the same identifications? Or should I think of it as "obvious" and just proceed forward? Any help, including a good reference that has a more detailed explanation on this topic, is welcome. Thanks in advance.

$\endgroup$

1 Answer 1

0
$\begingroup$

I am still not convinced that this is the best way to explain, but at least I came up with one way to explain this; hopefully, someone will find my explanation useful.

Suppose that $e_1,e_2,\dots,e_k$ are the edges of $P$. For each $i=1,\dots ,n$, let $\gamma _i(t)$ denote the point on $e_i$ corresponding to the parameter $t\in[0,1]$ (respecting the orientation given by the presentation). The equivalence relation on $P$ corresponding to the quotient map $q$ is generated by the relation $$R=\left\{(\gamma _i(t),\gamma_j(t))\mid e_i \text{ and } e_j \text{ are labeled with the same letter}, t\in[0,1]\right\}.$$ Let $R'$ denote the relation on $P'$ that is defined in the similar way. To check that $q$ and $q'\circ f$ make the same identifications, we just need to observe that the following statements are true for $x,y\in P$:

  • If $(x,y)\in R$, then $q'(f(x))=q'(f(y))$; and
  • If $(f(x),f(y))\in R'$, then $q(x)=q(y)$.

These statements are not difficult to prove, at least not as hard as to directly prove that $q$ and $q'\circ f$ make the same identifications by displaying the corresponding equivalence relations explicitly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.