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?