Let $X$ be a finite CW-complex and $\varphi:X\rightarrow X$ be a map. I want to prove that for the Lefschetz – number we have $\Delta(\varphi):=\underset{n\in\mathbb{N}}{\sum}(-1)^k\cdot\text{tr}(\varphi_\star:H_k(X)\rightarrow H_k(X))=\underset{n\in\mathbb{N}}{\sum}(-1)^k\cdot\text{tr}(\varphi^\star:H^k(X)\rightarrow H^k(X))$ and I would need some help with the argumentation.
So here is what I have:
We have the following commutative diagramm by the universal coeff. theorem
$\require{AMScd}$
\begin{CD}
0 @>>> \text{Ext}^1_\mathbb{Z}(H_{k-1}(X);\mathbb{Z}) @>>> H^{k}(X) @>>> \text{Hom}_\mathbb{Z}(H_{k}(X);\mathbb{Z})\rightarrow 0\\
@. @V ?VV @VV \varphi^\star V @VV g\mapsto g\ \circ\ \varphi_\star V @.\\
0 @>>> \text{Ext}^1_\mathbb{Z}(H_{k-1}(X);\mathbb{Z}) @>>> H^{k}(X) @>>> \text{Hom}_\mathbb{Z}(H_{k}(X);\mathbb{Z})\rightarrow 0\\
\end{CD}
The horizontal sequences are exact by the u.c.t. hence $\text{tr}(?)+\text{tr}(g\mapsto g\ \circ\ \varphi_\star)=\text{tr}(\varphi^\star)$ and $\text{tr}(g\mapsto g\ \circ\ \varphi_\star)=\text{tr}(\varphi_\star)$. If the homology groups are free abelian groups the Ext parts vanish and we are done.
If we have torsion in homology we know that in the definition of trace we only consider the free part of the groups. And now I would need some property of Ext I guess, e.g. it being torsion only for some reason..
If I didn't crash it somewhere else…
Thanks in advance!
Best Answer
Ext is always torsion in this context. Three extremely useful properties of Ext for Abelian groups (see for instance Hatcher chapter 3.1) that can be computed directly from taking free resolutions and quickly imply this are that for any Abelian group $G$: