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.).

enter image description here

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?


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:

enter image description here

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.

  • $\begingroup$ How are you attaching the $S^1$ which is the boundary of your 2-cell to your 1-cell? $\endgroup$ – Ian Coley Dec 19 '14 at 6:25
  • $\begingroup$ 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. $\endgroup$ – a student Dec 19 '14 at 7:02
  • $\begingroup$ 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? $\endgroup$ – Jack Davies Dec 19 '14 at 12:06
  • $\begingroup$ @JackDavies There is only one two cell in my question. $\endgroup$ – 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.

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  • $\begingroup$ Thank you, this resolves ${1\over 2}$ of my problem. Please, can you tell me the degree of your attaching map $\partial_2$? $\endgroup$ – a student Dec 20 '14 at 0:58
  • $\begingroup$ Is it $1$ because $\partial_2$ wraps $S^1=\partial D^2$ once around the circle with boundary $e^1_1 + e^2_1$? $\endgroup$ – a student Dec 20 '14 at 1:00
  • $\begingroup$ It's $\pm 1$, depending on orientations. With the "usual" orientations, it will be $+1$. $\endgroup$ – Ted Shifrin Dec 20 '14 at 1:01
  • $\begingroup$ Thank you. I'm not so clear on that. Say we put an orientation on the two edges and depict it by arrows. Do both arrows have to point in the same direction? (either both clockwise or both CCW?) $\endgroup$ – a student Dec 20 '14 at 1:08
  • $\begingroup$ 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. $\endgroup$ – Ted Shifrin Dec 20 '14 at 1:12

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