Euler Characteristic and cellular homology of $\ell^{th}-$space suspension.

Question

I am learning the Euler characteristic in terms of cellular homology and number of cells. Namely, for a space $X$, we have $$\sum_{j}(-1)^{j}c_{j}(X)=\chi(X)=\sum_{j}(-1)^{j}rank(H_{j}(X)),$$ where $c_{j}(X)$ is the number of $j$-cells in $X$.

The only note here, https://pages.uoregon.edu/ddugger/hw634-1.pdf, the Q5 of it, seems to claim that there is a way to connect $\chi(\Sigma^{\ell}X)$ with $\chi(X)$, where $\Sigma^{\ell}X$ is the $\ell^{th}$ suspension of $X$.

I tried to prove it but failed: since we have $$\chi(\Sigma^{\ell}Y)=\sum_{j}(-1)^{j}rank(H_{j}(\Sigma^{\ell}Y))$$ the only thing we need to do is to compute $H_{j}(\Sigma^{\ell}Y)$.

To do so, the only thing we can think about is the reduced homology: $$\overline{H}_{j}(\Sigma^{\ell}Y)=\left\{ \begin{array}{ll} H_{j}(\Sigma^{\ell}Y),\ \ \ \text{for}\ j>0;\\ \mathbb{Z}^{\{\#\ \text{of path components} - 1\}},\ \ \ \text{for}\ j=0. \end{array} \right.$$

But by the well-known Suspension Isomorphism, we have $$\overline{H}_{j}(\Sigma^{\ell}Y)=\overline{H}_{j-\ell}(Y).$$

But then what should I do next? Or is there another easier way to compute?

Thank you!

Answer 1

As Connor Malin suggests we could define a "reduced Euler Characteristic"

$$\tilde{\chi}(X) = \sum_i (-1)^irank(\tilde{H}_i(X)) $$

and then notice two things:

1) $\chi(X) = 1 + \tilde{\chi}(X)$,

2) by the suspension isomorphism $\tilde{\chi}(X) = -\tilde{\chi}(\Sigma X)$.

Putting these together we have

$$\begin{align}\chi(\Sigma X) &= 1 + \tilde{\chi}(\Sigma X) \\ &= 1 - \tilde{\chi}(X) \\ &= 1 - (\chi(X) - 1) \\ &= 2 - \chi(X).\end{align}$$

It follows that $\chi (\Sigma^{2k}X) = \chi(X)$ and $\chi(\Sigma^{2k+1}X) = 2 - \chi(X)$. As a "sanity check" recall that $\chi(S^{2k}) = 2$ and $\chi(S^{2k+1}) = 0$.