I'm studying for a qualifying exam in topology. Something that I struggled with during my course on homology theory was computing the signs of the cellular boundary maps; it wasn't until my course on cohomology theory that I realised the sign doesn't actually matter for computing homology (at least when the cellular chain groups have rank $1$). I also really struggled with understanding local degree, but I thought I understood it better after reading the section on degrees in Bredon's Topology and Geometry, which gives a result involving coordinate systems on $S^n$. However, I went to try Problem IV.10.2

Consider the space $X$ which is the union of the unit sphere $S^2$ in $\mathbb R^3$ and the line segment between the north and south poles. Give it a CW-structure and compute its homology,

and now I'm questioning my understanding again.

I gave $X$ a CW structure with two $0$-cells $e^0_1, e^0_2$ (the north and south pole), three $1$-cells $e^1_1, e^1_2, e^1_3$ (all with attaching map $-1 \mapsto e^0_1, 1 \mapsto e^0_2$), and two $2$-cells $e^2_1, e^2_2$ (both with attaching map $$e^{2 \pi i t} \mapsto \begin{cases} 4 t - 1 \in e^1_1 &\text{if $t \in [0, 1 / 2]$,}\\ -4 (t - 1 / 2) + 1 \in e^1_2 &\text{if $t \in [1 / 2, 1]$).} \end{cases} $$ Without considering signs, I was able to compute $$\partial e^1_i = \pm e^0_1 \pm e^0_2, \quad \partial e^2_i = \pm e^1_1 \pm e^1_2,$$ so that $$\partial_1 = \begin{bmatrix} \pm1 &\pm1 &\pm1\\ \pm1 &\pm1 &\pm1 \end{bmatrix} , \quad \partial_2 = \begin{bmatrix} \pm1 &\pm1\\ \pm1 &\pm1\\ 0 &0 \end{bmatrix} ,$$ and it seems to me that at least the relative signs in each row or column at least are important to determine the images and kernels of these homomorphisms, and I'm struggling to determine these.

To add some details, I know the cellular boundary formula as $\partial e^1_i = \sum_j d^1_{i j} e^0_j, \partial e^2_i = \sum_j d^2_{i j} e^2_j$, where $d^1_{i 1}$ is the degree of the map $$ \begin{array}{ccccccc} \partial D^1 &\to &X^{(0)} &\to &{X^{(0)} \amalg *} / e^0_2 \sim *\\ -1 &\mapsto &e^0_1 &\mapsto &e^0_1\\ 1 &\mapsto &e^0_2 &\mapsto &*, \end{array} $$ $d^1_2$ is the degree of the map $$ \begin{array}{ccccccc} \partial D^1 &\to &X^{(0)} &\to &{X^{(0)} \amalg *} / e^0_1 \sim *\\ -1 &\mapsto &e^0_1 &\mapsto &*\\ 1 &\mapsto &e^0_2 &\mapsto &e^0_2, \end{array} $$ $d^2_{i 1}$ is the degree of $$ \begin{array}{ccccccc} \partial D^2 &\to &X^{(1)} &\to &X^{(1)} / (X^{(1)} - \mathring e^1_1)\\ e^{2 \pi i t} &\mapsto &4 t - 1 &\mapsto &4 t - 1\\ e^{2 \pi i t} &\mapsto &-4(t - 1 / 2) + 1 &\mapsto &*, \end{array} $$ $d^1_{i 2}$ is the degree of $$ \begin{array}{ccccccc} \partial D^2 &\to &X^{(1)} &\to &X^{(1)} / (X^{(1)} - \mathring e^1_2)\\ e^{2 \pi i t} &\mapsto &4 t - 1 &\mapsto &*\\ e^{2 \pi i t} &\mapsto &-4(t - 1 / 2) + 1 &\mapsto &-4(t - 1 / 2) + 1, \end{array} $$ and $d^2_{i 3}$ is the degree of $$ \begin{array}{ccccccc} \partial D^2 &\to &X^{(1)} &\to &X^{(1)} / (X^{(1)} - \mathring e^1_3)\\ e^{2 \pi i t} &\mapsto &4 t - 1 &\mapsto &*\\ e^{2 \pi i t} &\mapsto &-4(t - 1 / 2) + 1 &\mapsto &*. \end{array} $$ To make sense of "degree", we have to choose a homeomorphism with $S^n$ for each of the target spaces, which choice determines the sign. However, since the target spaces are different, it doesn't seem like the choices are related, and any combination of homeomorphisms should give any combination of signs for the degrees.

  • $\begingroup$ The formula $\partial e_i^1 = e_2^0 - e_1^0$ should hopefully be intuitively obvious. The signs are for precisely the same reason as on the right hand side of the fundamental theorem of calculus $\int_a^b f(x)\ dx = F(b) - F(a)$ (and its upgraded version for line integrals). The attaching map for your $2$-cells seems to have some typos (I would assume the second line should define the map for $t \in [1/2,1]$), and it would be better for you to fix that before I attempt to explain anything. $\endgroup$ Commented Jun 11, 2023 at 6:00
  • $\begingroup$ @TabesBridges you're right about the attaching map; I fixed it. Also, yes, it is intuitively obvious, and I think I understand that one much better than the general case because Bredon's description of $X^{(0)} / \varnothing$ comes with a canonical base point, and w.r.t. that base point, one of the maps whose degrees appear in the boundary formula is the identity, and the other is the antipodal map. $\endgroup$ Commented Jun 11, 2023 at 14:47

1 Answer 1


First of all, note that the space $X$ is homotopy equivalent to $S^2 \vee S^1$ (imagine sliding the endpoints of the line segment together) so we quickly find that $\tilde{H}(X)\cong\tilde{H}(S^2)\oplus\tilde{H}(S^1)$ (Corollary 2.25 in Hatcher) which is a much faster way to solve the problem.

But to answer your actual question: luckily for attaching 1-cells and 2-cells, the signs of the cellular boundary map have an intuitive interpretation. Using your notation, in the first case, $\partial e^1=e^0_2-e^0_1$ when we attach $e^1$ via $-1\mapsto e^0_1$ and $1\mapsto e^0_2$.

The intuition of the second case is less exact, but when we attach a 2-cell, we can imagine the path/loop that the boundary circle of the 2-cell takes through the 1-cells in the 1-skeleton. If it goes through a 1-cell in the positive direction (i.e. from -1 to 1 using your notation), then that 1-cell is a positive term in $\partial e^2$, and if it goes in the opposite direction, then that 1-cell is a negative term. (If you want a bit more rigor, this is based on the fact that reflection on $S^1$ has degree $-1$ and reflection on $S^1$ is the same as a loop on $S^1$ going in the opposite direction).

So in your example (I am assuming the second mention of "if $t\in[0,1/2]$" is really "if $t\in[1/2,1]$") as we go from $t=0$ to $t=1$, we see that we first go along $e^1_1$ in the positive direction and then along $e^1_2$ in the negative direction (as $t=1/2$ gives us $1$ and $t=1$ gives us $-1$ in $e^1_2$). Thus $\partial e^2_i = e^1_1-e^1_2$ for each $i$. With this you can calculate that $H^{cell}_k(X)=\mathbb Z$ for $k=0,1,2$ and $0$ otherwise (as we know it should from my first paragraph).

For attaching $k$-cells with $k\geq3$, I don't know of anything simple that can help you calculate signs for cellular homology.

*After reading your clarifying edit, here is what I have to say: when you ask for a homeomorphism $X^{n-1}/(X^{n-1}-\mathring e^{n-1}_i)\cong S^{n-1}$, you are doing this with an identification of $e^{n-1}_i$ to $D^{n-1}$ already defined. So really we are asking for homeomorphisms $X^{n-1}/(X^{n-1}-\mathring e^{n-1}_i)\cong D^{n-1}/\partial S^{n-2}\cong S^{n-1}$. Here, the first homeomorphism is defined directly from how we identified $e^{n-1}_i$ with $D^{n-1}$, and the second homeomorphism is the same for any $n-1$-cell so we can make a canonical choice. Thus, the degree of each of your maps really does depend on how you orient each $k$-cell i.e. how you identify it with $D^k$.

  • $\begingroup$ That's more or less what I was thinking, but what determines the orientation on the lower-dimensional cells? Presumably all the orientations need to be consistent somehow, but I don't really know in what way that should be the case. $\endgroup$ Commented Jun 11, 2023 at 14:53
  • $\begingroup$ I'm not exactly sure what you're asking, but 0-cells don't have any orientation and 1-cells are oriented based on how you define the CW-complex i.e. which is the -1 end and which is the +1 end (and you can imagine an arrow pointing from -1 to +1 intuitively). As long as you're consistent, it doesn't matter how these are oriented in your construction of the CW-structure i.e. different choices of CW-structures for the same space will give isomorphic cellular homology groups. Please let me know if I didn't understand your question. $\endgroup$
    – Adam Z
    Commented Jun 11, 2023 at 17:12
  • $\begingroup$ (As an aside, there is a notion of orientation on $0$-cells by simply assigning to a $0$-cell a choice of positive or negative). So, you're saying that the orientation comes from a fixed orientation (say, it's orientation at a subset of Euclidean space) on the balls of each dimension? $\endgroup$ Commented Jun 11, 2023 at 18:49
  • $\begingroup$ And that the sign comes from the the induced orientation of a sphere as the boundary of a cell? $\endgroup$ Commented Jun 11, 2023 at 19:18
  • $\begingroup$ Yes that is the right idea. And in terms of how to pick a fixed orientation of a sphere, the notion I am familiar with is picking certain generators for the nth reduced homology group of each n-sphere. On the 0-sphere, the usual choice is the chain s(1)-s(-1) where s(p) is the 0-simplex mapping to the point p. Then using any standard induction proof to calculate the homology groups of spheres, we can then inductively define a choice of generators for the rest of the spheres. There are surely more direct ways of defining orientation, but this is what I am familiar with. $\endgroup$
    – Adam Z
    Commented Jun 11, 2023 at 19:55

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