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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.

Dmitry
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  • Do you know the proof of multiplicativity of $\chi$ under finite covering maps? Did you try to use that proof in your situation? As for the other question, it has negative answer. Try to find a counter example. – Moishe Kohan Dec 12 '22 at 11:21
  • @MoisheKohan Following the answer, let $f^{-1}(x)={y_1(x),\ldots,y_m(x)}$ and define $m$ "lifts" by $h_{\alpha i}^k\colon D_\alpha^n\ni a\mapsto y_i(\phi_\alpha^k(a))\in Y$, where $\phi_\alpha^k\colon D_\alpha^k\to X$ are characteristic maps of CW-complex $X$. It is seen that $f\circ h_{\alpha i}^k=\phi_\alpha^k$. If dimension of $X$ is $n$, how one can show that dimension of $Y$ is also $n$? – Dmitry Dec 21 '22 at 12:59
  • @MoisheKohan About the counterexample: I can't even produce any example of a piecewise linear map with constant fiber between polyhedra. I've tried to construct a map from triangle to segment but I don't know how to deal with angles of the triangle. – Dmitry Dec 21 '22 at 13:03

1 Answers1

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  1. 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.

  2. 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.)

Moishe Kohan
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