I am very inexperienced with Algebraic Topology but I am trying to put together a proof for the Lefschetz Fixed Point Theorem, that I can understand, for a presentation in class.
I am following Prasolov's "Elements of Homology Theory" as closely as I can. And for the record, I am only meant to present this proof to the class, not create it myself. I just want to understand what is going on to the best of my ability, so I have rewritten some things to that end.
I will post what I have so far below but I am wondering if someone can help me finish off this proof using the Hopf Trace Lemma that I have in the beginning? And if anyone spots any mistakes or other ways that I might finish off the proof that would be simple enough for a very lost student to try and wrap his head around, that would be much appreciated.
Before getting to the proof of the theorem, we are going to need the following lemma.
Lemma (Hopf Trace):
Let $X$ be a finite simplicial complex and let $\phi:C_n(X)\to C_n(X)$ be a chain map. Then \begin{align*} \sum_{n\ge0}(-1)^n\text{tr}\big(\phi:C_n(X)\to C_n(X)\big)=\sum_{n\ge0}(-1)^n\text{tr}\big(\phi_*:H_n\big(X;\mathbb{R}\big)\to H_n\big(X;\mathbb{R}\big)\big). \end{align*}
Note: This formulation of the statement comes from Munkres (1984), ``Elements of Algebraic Topology'' and we have the used here the fact that any homology groups with real coefficients have the trivial torsion subgroup.
Theorem (Lefschetz Fixed-Point Theorem):
Let $X$ be a finite simplicial complex. If $f:X\to X$ is a map with Lefschetz number $\Lambda(f)\ne0$, then $f$ has a fixed point.
Proof
Suppose to the contrary that $f$ has no fixed points. Then, to prove the theorem, we must show that in fact $\Lambda(f)=0$.
Since $X$ is a finite simplicial complex, it is compact. To see this, note that since the simplicial complex $X$ is finite, it is constructed by finitely many simplices arranged in a finite manner. Since there are only finitely many simplices in $X$, we can construct an open cover of $X$ by taking the union of open neighborhoods of each vertex of each simplex. This open cover is finite because there are only finitely many vertices and simplices in $X$.
Now let's consider the distances between the points $x\in|X|$ and their images $f(x)$. Recall the assumption that $f$ has no fixed points, meaning that there does not exist an $x\in|X|$ such that $x=f(x)$. So, for each point $x\in|X|$, we may take an open ball centered at $x$ but excluding the point $f(x)$. The union of all of these open sets is a cover for $X$ and since $X$ is compact, we may take a finite subcover.
Next, since $X$ is finite, there must exist some subdivision $X'$ of $X$ such that the maximum diameter of any simplex in $X'$ is less than or equal to the minimum distance between the centers of the open sets in this finite subcover. Furthermore, we may in fact select this subdivision $X'$ to be such that for all $x\in|X|$: the distance between $x$ and $f(x)$ is larger than the maximum diameter of any simplex from $X'$.
Now let $X''$ be a further subdivision of $X'$ and let $\phi:X''\to X'$ be a simplicial approximation of $f$. Consider an arbitrary simplex $\Delta'' \subset X''$ and let $x''$ be any point within the interior of $\Delta''$. Then, the point $\phi(x'')$ belongs to the only simplex of $X'$ whose interior contains $f(x'')$; but recall that we've ensured the distance between $x''$ and $f(x'')$ is greater than the maximum diameter of any simplex in $X'$. Thus, since $X''$ is a subdivision of $X'$, the simplex in $X'$ containing both $f(x'')$ and $\phi(x'')$ must be disjoint from the simplex in $X''$ containing $x''$. Finally, since $x''\in\Delta''$ was arbitrary, we have that for any simplex $\Delta''\subset K'$, the sets $\Delta''$ and $\phi(\Delta'')$ are disjoint.
*** This is where I am not sure how to proceed to finish things off ***