# Attaching a 2-cell by a word

In Example 2.36 on Pg 141 of Hatcher's Algebraic Topology, he writes: ... one 2-cell attached by the product of commutators $[a_1,b_1] \ldots$

Can someone please explain to me what is meant by attaching a 2-cell by a word. I am assuming it means that the attaching map can be given by a word in some free group. But, I cannot make sense of this.

Thank you for your time.

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You start with a wedge of $2g$ oriented circles, labeled by $a_i,b_i$. Then each letter in the word corresponds to gluing part of the boundary of the 2-cell to that $1$-cell, in the orientation prescribed by whether you have the letter or its inverse. You get the complete word by travelling around the boundary of the $2$-cell.

Also, look at the picture on page 5 of Hatcher. This explains the construction in detail. For example, the genus 3 surface pictured there has word $[a,b][c,d][e,f]$ as you travel around the boundary of the $2$-cell.

Edit: Here is an actual formua in the case of the torus. We have two circles identified at a point. Let them be parameterized by $\theta_1$ and $\theta_2$ respectively where $\theta_1,\theta_2\colon[0,2\pi]\to S^1$, with basepoint $\theta_1(0)=\theta_2(0)$. Okay, now think of the $2$-cell as a unit square $[0,1]\times[0,1]$. The function from the boundary of the square to the wedge of two circles is given by

$$(0,y)\mapsto \theta_2(2\pi y),\,\,\,\, (1,y)\mapsto \theta_2(2\pi y)$$ $$(x,0)\mapsto\theta_1(2\pi x),\,\,\,\,\, (x,1)\mapsto\theta_1(2\pi x)$$

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Even with the picture on Pg 5, it seems almost impossible to visualize. Even in the 1-g torus case. After I paste the bottom edge of the square against $a$, I cannot see a way to paste the two verticle edges of the square (now identified) along the circle $b$. Or am I going about this wrong? – doofus Nov 2 '11 at 15:57
Also I can figure out a fuction from a unit square to the torus (defined as a surface of revolution) and can show it gives the required homeorphism between the quotient construction and the torus. But, in this case I have no idea of how to convert this word to a formula. – doofus Nov 2 '11 at 16:02
@doofus: In the torus picture, first glue the two sides named $b$ together to get a cylinder, now bend the cylinder around so that the two circles are glued together. Then you get a torus. In the end, the $a$ and $b$ edges actually glue up as circles identified at a point. So, in retrospect, you could have started with them glued up that way. – Grumpy Parsnip Nov 2 '11 at 16:20
I will edit to make clearer what the "formula" is. – Grumpy Parsnip Nov 2 '11 at 16:22
In your formula, shouldn't $(x,1) \mapsto -\theta_1(2\pi x)$ and similarly for $(0,y)$? Also, how can we extend this to a general point inside the square? Thanks a lot for the help. – doofus Nov 2 '11 at 16:46

I think you just need more familiarity with the general notion of adjunction spaces of the form $B \cup _fX$ where $f: A \to B$ and $A$ is a closed subspace of $B$, with inclusion $i: A \to B$. The most important property of this is that the square of maps

$$\begin{matrix} A & \to & B \\ \downarrow&& \downarrow\\ X &\to& B \cup _fX \end{matrix}$$ is a pushout, i.e. you can construct in a unique way continuous functions $B \cup _fX \to Y$ by giving a pair of functions $g: X \to Y, h: B \to Y$ such that $gi= hf$. In this way you do not need to visualise $B \cup_fX$, though a few simple examples are useful, since this is the way we use them. See my book "Topology and groupoids".

In the case $A=S^1$, $X= E^2$ then a map $f: A \to B$ represents an element of $\pi_1(B)$, and if this is a free group then its elements can be written as words in the generators.

October 5: More can be explained through an example. Suppose $K$ is the Klein Bottle obtained in the usual way as an identification of a square $\sigma$ with sides in order $a,b,-a,b$. We would like to write $$\partial \sigma = a+b-a+b.$$ The RHS is clearly an element of the free group on two generators $a,b$ which correspond to the two circles of the $1$-skeleton $K^1$ of the Klein Bottle. But what is the LHS, and what is $\partial$?

It turns out that this is $\partial : \pi_2(K,K^1,v) \to \pi_1(K^1,v)$ , that this has the structure of crossed module, and the second relative homotopy group is the free crossed module on one generator $\sigma$.

This result is not so easy to prove, but it is the start of lifting the nonabelian ideas of the fundamental group and groupoid to dimension $2$. See the book (with free pdf) advertised here.

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