# Degree of this attaching map — or how to define this attaching map?

Consider the cell complex consisting of two zero cells $e_0^1, e_0^2$ connected by two 1 cells $e_1^1,e_1^2$ with one 2 cell $e_2$ in the middle (Picture: Imagine $S^1$ with one $0$-cell at the north pole and the other at the south pole. Glue the disk into this circle.). Let $f_1: S^1 \to S^1$, $f_2: S^1 \to S^1$ be the maps that attach the two cell to the two $1$-cells. I am trying to calculate the degree of $f_1,f_2$.

Here the degree is described as " the degree of a continuous mapping between two compact oriented manifolds of the same dimension is a number that represents the number of times that the domain manifold wraps around the range manifold under the mapping."

I understand that. If I apply it to the $2$-cell in my example then half of the $2$-cell wraps once around one of the $1$-cells while the other half wraps around the other $1$-cell.

Therefore we should have $\deg(f_1) = \deg(f_2) = {1\over 2}$. The problem is, in the definition the degree is an integer. How do I calculate the degree of this attaching map correctly?

Edit

Maybe my problem is that I don't know how to define the attaching map for this example so that it is one attaching map $S^1\to S^1$ rather than two maps. After all, the degree is define for one map $S^n \to S^n$. But I don't see how I can calculate the degree if I define it using one map.

Edit 2

With orientation as assumed in the answer by Ted Shifrin: Edit 3

In this question I am asking about degrees of attaching maps $\chi_n$ as in the definition of cellular homology on Wikipedia. I am not asking about boundary maps.

• How are you attaching the $S^1$ which is the boundary of your 2-cell to your 1-cell? – Ian Coley Dec 19 '14 at 6:25
• I'm attaching it to 2 2-cells: half to each of it. I tried to explain it by a picture in my first sentence. Sorry that it's not so clear. – a student Dec 19 '14 at 7:02
• So you're attaching one 2-cell to one of your 1-cells, and your second 2-cell to the remaining 1-cell. Or are you attaching each 2-cell to all of your previous 1-skeleton? Your 1-skeleton is also homeomorphic to $S^1$ so why don't we start with that? – Jack Davies Dec 19 '14 at 12:06
• @JackDavies There is only one two cell in my question. – a student Dec 19 '14 at 23:19

You're misinterpreting the boundary map. There is just one circle, formed by the union of the two $1$-cells. (The degree is defined from the boundary of each $2$-cell to each circle in the $1$-skeleton.) So the boundary of the $2$-cell is the sum $e_1^1+e_1^2$. Note that you have $C_2 \cong \Bbb Z$, $C_1 \cong \Bbb Z\oplus\Bbb Z$, and $C_0 \cong \Bbb Z\oplus \Bbb Z$. We have \begin{align*} \partial_2\colon C_2\to C_1\,, &\quad \partial_2(e_2) = e_1^1+e_1^2, \\ \partial_1\colon C_1\to C_0\,, &\quad \partial_1(e_1^1) = e_0^2-e_0^1, \partial_1(e_1^2)= e_0^1-e_0^2. \end{align*} You can check that this gives the homology $H_2 \cong 0$, $H_1\cong 0$, $H_0\cong\Bbb Z$, as it should.
• Thank you, this resolves ${1\over 2}$ of my problem. Please, can you tell me the degree of your attaching map $\partial_2$? – a student Dec 20 '14 at 0:58
• Is it $1$ because $\partial_2$ wraps $S^1=\partial D^2$ once around the circle with boundary $e^1_1 + e^2_1$? – a student Dec 20 '14 at 1:00
• It's $\pm 1$, depending on orientations. With the "usual" orientations, it will be $+1$. – Ted Shifrin Dec 20 '14 at 1:01
• They can be oriented any way you want, but the formulas I wrote above assumed that $e_1^1$ was on the left, heading down, $e_1^2$ on the right heading up. So their sum is a circle oriented CCW. – Ted Shifrin Dec 20 '14 at 1:12