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Let $f\cn Y\to X$ be a piecewise-linear map between two compact polyhedra. Suppose that the preimage of each point consists of precisely $m$ points. Prove that $\chi(Y)=m\chi(X)$. Must $f$ be a covering?
I've seen that surjective local homeomorphism between Hausdorff spaces with constant and compact fibers is a covering map. So if $f$ is local homeomorphism then it is a covering. Consequently, if $f$ is open and locally injective then it is a covering. However, I don't know how to prove or disprove it.
The following is also true. If $X$ is a finite CW-complex and if $Y\to X$ is a $m$-sheeted covering then $Y$ is a finite CW-complex and $\chi(Y)=m\chi(X)$. Therefore if $f$ is indeed a covering then we're done.
Any help would be appreciated.
Here is what's true: Suppose that $X, Y$ are finite simplicial complexes with geometric realizations $|X|, |Y|$, and $f: X\to Y$ is a simplicial map such that the corresponding map (also denoted $f$) of geometric realizations. Assume that there exists $m<\infty$ such that $f$ satisfies $$ card(f^{-1}(y))=m, \forall y\in |Y|. $$ Then $\chi(X)=m\chi(Y)$. The proof is direct counting, exactly the same as for the covering maps.
Here is what's false: Each map $f$ in (1) is a covering map. This is already false in the case of 1-dimensional simplicial complexes.
i. Consider $Z=[-1,0]\cup [0,1]$, $X=Z\cup \{2\}$ and $Y=[0,1]$ with their natural simplicial complex structures. Take the simplicial map $f: X\to Y$ such that $f(x)=|x|, x\in |Z|$ and $f(2)=0$. Then $card(f^{-1}(y))=2$ for each $y\in |Y|$, but $f$, of course, is not a covering map.
ii. If you do not like the fact that $X$ is not connected, take $X$ to be $S^1$ triangulated to have 6 edges and $Y$ to be $S^1$ triangulated to have 3 edges. You will find a simplicial map $f: X\to Y$ such that the preimage of each point in $|Y|$ has cardinality 2, but $f$ is not a covering map. (If you do not see the map immediately, it is a finite search problem to find one.)