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Let $X$ be graph and $p:E\rightarrow X $ be covering map. It is followed by Theorem $83.4$, $E$ is graph. Now, assume that $v\in V(X)$ is a vertex with $deg(v)<\infty$ and $w\in p^{-1}(v)$.

Can we conclude that $deg(v)=deg(w)$?

I've tried this:

Suppose that $U$ is the open neighborhood such that evenly covered by $p$. So $p^{-1}(U)$ is disjoint union of open nbhd $V_{\alpha}$ in $E$ such that $V_{\alpha}$ is homeomorphism to $U$. Note that topology of $X$ is coherent with each of edge. Suppose that $U\cap A_i$ is not empty where $A_i$ is edge for $i=1,\ldots,deg(v)$. Then $U=\cup_{i=1}^{i=deg(v)} (U\cap A_i)$. Since every open set $V_{\alpha}$ is homeomorphism to $U$, we can conclude that $deg(v)=deg(w)$ where $p(w)=v$.

user108209
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1 Answers1

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I think you have the right idea (degree is topologically a local notion), but your arguments are a bit confusing. In particular it's not clear why it follows that $U = \bigcup (U \cap A_i)$, and it's not necessary to assume that $U \cap A_i$ is nonempty because that will always be true...

Here is a more rigorous argument: there is a neighborhood $W \subset E$ of $w$ that is mapped homeomorphically to a neighborhood $p(W) = V \subset X$ of $v$. By making this neighborhood smaller, we can assume that it is contained in the union of the edges touching $v$. Then the degree of $v$ is the number of connected components of $V \setminus \{v\}$. Since $p : W \to V$ is a homeomorphism, so is $W \setminus \{w\} \to V \setminus \{v\}$, so they have the same number of connected components, and therefore $\operatorname{deg}(w) = \operatorname{deg}(v)$.

Najib Idrissi
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