# When a quotient map of topological graph is open?

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!

• 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... – Misha Lavrov Jan 31 at 21:32
• Thank you very much. Now I also understand the topology better. I have updated the question, would be happy if you can check it. – User3231 Feb 1 at 3:13

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.

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