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If $X$ is a finite simplicial complex and $f:X\rightarrow X$ is a simplicial homeomorphism, show that the Lefschetz number $\tau(f)$ equals the Euler characteristic of the set of fixed points of $f$.

My progress: We know the set of fixed points $X_f$ of $f$ can be represented as a simplicial complex by barycentric subdivision on $X$. Then, the faces of $X_f$ are generated by the fixed vertices of each maximal face and the barycentric point of such a face. I've created some examples to verify the problem statement, but this creates no intuition on how to proceed. I realize that there are symmetries that determine what faces could be mapped into what other faces. Any hints on how to proceed? Especially how to relate the trace of $f_*:H_n(X)\rightarrow H_n(X)$ to the rank of $H_n(X_f)$ or the number of cells of $X_f$?

EDIT: I've tried to work on this for a bit, but the terminology and conclusions that user8268 makes are too complicated for what I know, so I'll list what I'm working with.

We have the short exact sequences $0\rightarrow Z_n\rightarrow C_n\rightarrow B_{n-1}\rightarrow 0$ and $0\rightarrow B_n\rightarrow Z_n\rightarrow H_n\rightarrow 0$.

Then, we can write $C_n=B_{n-1}\oplus Z_n=H_n\oplus B_n\oplus B_{n-1}$, so I'm assuming this is what the graded vector space. Then, through some manipulation, the trace of the induced endomorphism on $H_n$ is equal to the trace of the map $\frac{ker\partial_n}{im\partial_{n+1}}f_*:H_n\oplus B_n\oplus B_{n-1}\rightarrow H_n$, but how does this matrix look like (this doesn't even look like a square matrix)?

Can someone offer a simpler explanation or an alternative approach?

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After barycentric subdivision, $f$ gives you a permutation of the $k$-dim simplexes, the simplexes which are fixed by this permutation are those which are in the fixes point set of $f$. Hence $f_*:C_k(X)\to C_k(X)$ is a permutation matrix, so its trace is the number of the fixed simplexes, i.e. $\dim C_k(X_f)$.

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I did have this in mind, too, but how do you get from the matrix involved in $C_k(X)\rightarrow C_k(X)$ to that of $H_k(X)\rightarrow H_k(X)$. I think that's where my confusion lies. – ergo May 4 '11 at 7:34
@ergo: Euler characteristics is $\sum_k (-1)^k\dim H_k(X)=\sum_k (-1)^k\dim C_k(X)$. – user8268 May 4 '11 at 8:12
I don't see how this relates to the trace of $f_*$. I want to show $\tau(f)=\chi(X_f)$ or after substitutions, this becomes $\sum (-1)^n\operatorname{Tr}(f_*: H_n(X)\rightarrow H_n(X) )=\sum (-1)^n\operatorname{Tr}(f_*: C_n(X)\rightarrow C_n(X) )$, which is the source of my comment. – ergo May 4 '11 at 8:42
$\sum(-1)^1Tr(f_*:C_k(X)\to C_k(X))=\sum(-1)^k Tr(f_*:H_k(X)\to H_k(X))$ has roughly the same proof as $\sum(-1)^1\dim C_k(X)=\sum(-1)^k \dim H_k(X)$: The complex $C_\bullet(X)$ is isomorphic (non-canonically) to the direct sum of $H_\bullet(X)$ (with zero differential) with the cone of a graded vector space $V_\bullet$ ($CV_k=V_{k+1}\oplus V_k$, $d:CV_k\to CV_{k-1}$ is the identity on $V_k$ and zero on $V_{k+1}$). The map $f_*$ commutes with the differential. Use that if $A\in End(V\oplus W)$ then $Tr(A)=Tr(A|_V)+Tr(A|_W)$ and watch the alternating sum killing pieces of $Tr(f_*)$. – user8268 May 4 '11 at 9:23

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