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I follow the definition of a topology on a graph, from wikipedia:

A graph is a topological space which arises from a usual graph $G=(E,V)$ by replacing vertices by points and each edge $e=xy\in E$ by a copy of the unit interval $ I=[0,1]$, where $0$ is identified with the point associated to $x$ and $1$ with the point associated to $y$. That is, as topological spaces, graphs are exactly the simplicial $1-$complexes and also exactly the one-dimensional CW complexes.

Thus, in particular, it bears the quotient topology of the set

$X_{0}\sqcup \bigsqcup_{e\in E}I_{e}$ under the quotient map used for gluing.

My question is when this quotient map is an open map? I have the impression that when the graph is locally finite it should be fine. But are there specific conditions making it open? Or, maybe it is always open? I would like to see counterexamples as well.

$\textbf{EDIT:}$ As was clarified in the comments, the answer is no. However, what I originally wanted to get was:

(1) If $A$ is a closed subset in the graph (w.r.t the quotient topology) and $\mathrm{int}(A)=\emptyset$, then $\mathrm{int}(f^{-1}(A))=\emptyset$.

Of course, if the map was open, I could get it, as for any $A$,

$f^{-1}(\mathrm{int}(A))=\mathrm{int}(f^{-1}(A))$.

But it's enough for me to get (1), and now I have the impression that it should be true, as the inverse image would at most contain more point from $X_0$. Is it true?

Thank you!

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  • $\begingroup$ If we take the open set in $X = X_{0}\sqcup \bigsqcup_{e\in E}I_{e}$ consisting of just one $I_e$, and if either endpoint of $e$ has degree $>1$, then as far as I can tell its image in $X /{\sim}$ won't be open... $\endgroup$ – Misha Lavrov Jan 31 at 21:32
  • $\begingroup$ Thank you very much. Now I also understand the topology better. I have updated the question, would be happy if you can check it. $\endgroup$ – User3231 Feb 1 at 3:13
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Unfortunately, the desired statement is false. The problem is that even if $\operatorname{int}(A) = \emptyset$, if $A$ contains any vertices of the graph topological space, then $f^{-1}(A)$ contains some elements of $X_0$. These are isolated points of $X_{0}\sqcup \bigsqcup_{e\in E}I_{e}$, and therefore they're automatically in the interior of $f^{-1}(A)$: if $x \in X_0 \cap f^{-1}(A)$, then $\{x\} \subseteq f^{-1}(A)$ is an open set containing $x$.

However, this is the only problem: if $\operatorname{int}(A) = \emptyset$, then $\operatorname{int}(f^{-1}(A)) \subseteq X_0$. To see this, suppose for the sake of contradiction that there is some $x \in \operatorname{int}(f^{-1}(A))$ such that $x \in I_e$. Then there is some open set $U \subseteq f^{-1}(A) \cap I_e$ containing $x$.

This set $U$ contains an open interval of $I_e$. But away from the endpoints of $I_e$, $f$ is a bijection between the interior of the interval $I_e$, and the interior of the edge. So $f$ will map this open interval of $I_e$ to an open interval of the edge, which is contained in $A$.

This contradicts the assumption that $\operatorname{int}(A) = \emptyset$. So, if $A$ has empty interior, then the interior of $f^{-1}(A)$ is contained entirely in $X_0$.


To avoid this, we can take a slightly different definition of the topological space: start with just $\bigsqcup_{e\in E}I_{e}$, and have the quotient map identify endpoints of the intervals that are supposed to represent the same vertex. Exception: if there are isolated vertices in the graph, we can take a set $X_0$ consisting of just those vertices, and have the quotient map leave those alone.

Now, if $\operatorname{int}(A) = \emptyset$, in particular $A$ contains no isolated vertices (those would be interior points of $A$) so $f^{-1}(A)$ is entirely contained in $\bigsqcup_{e\in E}I_{e}$, and the argument above proves that $\operatorname{int}(f^{-1}(A))$ is empty.

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  • $\begingroup$ Thank you! The alternative dedinition you gave solves for me the problem. $\endgroup$ – User3231 Feb 1 at 9:29

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