Consider a graph $X$ with two vertices $a, b$, three edges :

$e_1$ : edge from $a$ to $b$

$e_2, e_3$ : loops at $a$ and $b$, respectively.

I want to show that this graph is not a covering space for any other space $Y(\neq X)$. I saw some people said if $X \to Y$ is a covering map, then $Y$ must be a graph with a 3-valence single vertex, which is impossible.

However, it is not trivial to me that if $X \to Y$ is a covering map then $Y$ is a graph. Is this true?

Second, if $X \to Y$ is a covering map with $X, Y$ graphs, then should every vertex of $X$ map to a vertex of $Y$?

I am very confused. Am I missing something?

  • 1
    $\begingroup$ You're not missing anything; the first assertion is nontrivial and the second is false. Anyone who claimed this has such an easy solution is probably just thinking intuitively and not worrying about how to rigorously justify their geometric intuition. The idea certainly can be turned into a proof but how exactly to do so is not obvious. $\endgroup$ Aug 18, 2019 at 0:48
  • $\begingroup$ Are you comfortable with the theory of CW complexes? $\endgroup$
    – Elad
    Aug 18, 2019 at 7:23
  • $\begingroup$ @Elad I know some basic properties, for example, some of them in Lee's "Introduction to Topological Manifolds" $\endgroup$
    – user302934
    Aug 18, 2019 at 15:23
  • $\begingroup$ Of course there exist covering maps $p : X \to Y$ with $Y \ne X$. What you mean is that each such covering map is a homeomorphism.. $\endgroup$
    – Paul Frost
    Aug 18, 2019 at 15:29
  • $\begingroup$ @Paul Frost your right $\endgroup$
    – user302934
    Aug 18, 2019 at 15:35

2 Answers 2


I found that it is intuitively clear, but a rigorous proof is not easy. Here is my attempt. Perhaps it can be shortened, but I do not see that.

Let $p : X \to Y$ be a covering map which is not a homeomorphism. Since $X$ is path-connected, so is $Y$ and all fibers $p^{-1}(y)$ have the same cardinality. This cardinality cannot be $1$ because $1$-sheeted covering maps are homeomorphisms. We claim that $p$ is a $2$-sheeted covering map.

Let us first verify that $p^{-1}(p(a)) \subset \{ a, b\}$ and thus $p^{-1}(p(a)) = \{ a, b\}$ because fibers have more than one point. Hence all fibers have $2$ elements.

Assume that there exists $x \notin \{ a, b\}$ such that $p(x) = p(a)$. Take an open neighborhood $V$ of $p(a)$ which is evenly covered. Hence $p^{-1}(V)$ is the disjoint union of open subsets $U_\iota \subset X$ such that the restrictions $p_\iota : U_\iota \to V$ are homeomorphisms. Let $x \in U_{\iota_x}, a \in U_{\iota_a}$. Let $U$ be an open neighborhood of $x$ which is contained in $U_{\iota_x}$ and homeomorphic to an open interval. Then $U' = p_{\iota_a}^{-1}(p_{\iota_x}(U)$ is an open neighborhood of $a$ which is homeomorphic to an open interval. Hence $U' \setminus \{a\}$ has two components. This is a contradiction because $U' \setminus \{a\}$ must have at least three components.

Since fibers are finite and $X$ is compact Hausdorff, also $Y$ is compact Hausdorff (see https://math.stackexchange.com/q/3969891/).

Now let $Y' = Y \setminus \{p(a)\}$ and $X' = p^{-1}(Y') = X \setminus \{ a, b\}$. Then $p$ restricts to a $2$-sheeted covering map $p' : X' \to Y'$; clearly $Y'$ is Hausdorff. The space $X'$ has three path-components $X_1, X_2, X_3$ which are homeomorphic to $\mathbb R$. Hence $Y'$ has at most three path-components $Y_i$. Each $p(X_j)$ is path connected and thus contained in a unique $Y_{i(j)}$. We have $p(X_j) = Y_{i(j)}$. (Let $y \in Y_{i(j)}$. Pick $x \in X_j$ and choose a path $u$ in $Y_{i(j)}$ from $p(x)$ to $y$. It has a lift to a path $u'$ in $X$ starting at $x$. This path must be a path in $X_j$, thus $u'(1) \in X_j$ and $y = u(1) = p(u'(1)) \in p(X_j)$.)

Hence the restriction $p'_j : X_j \to Y_{i(j)}$ of $p'$ is a continuous surjection whose fibers have at most two points (it can easily be shown that it is a $1$- or $2$-sheeted covering map, but that is irrelevant).

No $Y_i$ can be compact. (If $Y_i$ would be compact, then $Y \setminus Y_i$ would be an open neighborhood of $p(a)$. Pick open neighborhoods $W_a, W_b$ of $a, b$ such that $p(W_a), p(W_b) \subset Y \setminus Y_i$. We have $p(X_j) \subset Y_i$ for some $j$. But $X_j$ contains a point $x_j \in W = W_a \cup W_b$, thus $p(x_j) \in Y \setminus Y_i$ which is a contradiction.)

Since $p'$ is a local homeomorphism, $Y'$ is locally Euclidean of dimension $1$. Moreover, $Y'$ has a countable base (take the image of a countable base of $X'$). Hence $Y'$ is a topological $1$-manifold with at most three components $Y_i$. Each $Y_i$ is either homeomorphic to the circle $S^1$ or to $\mathbb R$ (see {The only 1-manifolds are $\mathbb R$ and $S^1$). Since the $Y_i$ are not compact, they are homeomorphic to $\mathbb R$.

Let us identify $p'_j : X_j \to Y_{i(j)}$ (using homeomorphisms between $X_j, Y_{i(j)}$ and $\mathbb R$) with a map $g : \mathbb R \to \mathbb R$. Each $t \in \mathbb R$ has at most two preimages under $g$. We claim that $g$ (and hence also $p'_j$) is injective.

Assume that there are $x_1,x_2 \in \mathbb R$ such that $x_1 < x_2$ and $g(x_1) = g(x_2) = t$. The set $g([x_1,x_2])$ is a compact connected subset of $\mathbb R$, thus a closed interval $[r,s]$ with $r < s$ ($r = s$ would imply that $g^{-1}(r)$ is infinite).

Let us assume $t \in (r,s)$. Then we find $r', s' \in (x_1,x_2)$ such that $g(r') =r, g(s') = s$. Let $J$ be the closed interval with boundary points $r',s'$. By the intermediate value theorem $g(J) =[r,s]$, hence there is $x_3 \in J \subset (x_1,x_2)$ with $g(x_3) = t$ which means that $t$ has at least three preimages which is a contradiction.

Thus we must have $t=r$ or $t=s$. We only consider $t=r$, the other case is similar. Choose $s' \in (x_1,x_2)$ with $g(s') = s$. By the intermediate value theorem $g([x_1,s'] = g([s',x_2]) = [r,s]$. Hence each point of $[r,s)$ has at least two preimages in $[x_1,x_2]$. We cannot have $r \in g((x_2,\infty))$ (otherwise $r$ would have more than two preimages), thus $ g((x_2,\infty))$ is either contained in $(-\infty,r)$ or in $(r,\infty)$. If it were contained in $(r,\infty)$, then $g((x_2,\infty))$ would contain points of $(r,s)$ because $g(x_2) =r$. Each such point would have more than two preimages which is impossible. Thus $g((x_2,\infty)) \subset (-\infty,r)$. A similar argument shows that $g((-\infty,x_1)) \subset (-\infty,r)$.

We conclude that $g$ cannot be surjective which is a contradiction.

Knowing that the $p'_j$ are bijections, we now finish our proof.

If $Y'$ has one component, then each $y \in Y'$ would have three preimages.

If $Y'$ has two components, then two of the components of $X'$ are mapped onto a component $Y_1$ and the other component of $X'$ onto the second component $Y_2$. Hence the elements of $Y_2$ would have only one preimage.

If $Y'$ has three components, then we get three bijections $p'_j$ which implies that all elements of $Y'$ have only one preimage.

Therefore any covering map $p : X \to Y$ must be a homeomorphism.


Assume $X$ is a covering space of some space $Y \ne X$ and $p:X \to Y$ the covering map. Lets prove that $p(a)=p(b)$ where $a,b$ are the vertices of $X$. If $p^{-1}(\{p(a)\})=\{a\}$ then $X=Y$, else exist $a \ne x \in p^{-1}(\{p(a)\})$. If $x \ne b$ then there exist a neighborhood of $x$ homomorphic to $(0,1)$ but this is a contradiction. As small neighborhood of $a$ are not homeomorphic to $(0,1)$ and small enough neighborhoods of $a$ and of $x$ are homeomorphic to small enough neighborhood of $p(a)$. This shows that $p(a)=p(b)$.

Now look at the edge $(a,b)$ I claim the map $p_{\restriction (a,b)}$ is injective. If you assume the contrary you will get that there is a point in $Y$ that is not $p(a)$ that has a neighborhood not homeomorphic to $(0,1)$. We can do the same thing with $(a,a)$ and $(b,b)$ and conclude that inverse image of one $p((a,b)), p(a,a), p(b,b)$ has a fiber of cardinality 1.

I will elaborate on the proof that $p_{\restriction (a,b)}$ is injective. Assume it is not and let $x_2 \in (a,b)$ the first element such that there is $x_1< x_2$ that $p(x_1)=p(x_2)$. Now choose a small enough neighborhood $U$ of $p(x_1)$. such that $p^{-1}(U)=U \times F$. Remove the point $p(x_1)$ from $U$ and observe that $p_{\restriction (x_1-\varepsilon,x_1)\cup (x_1, x_1 +\varepsilon)\cup(x_2 -\varepsilon,x_2) }$ is injective. This means that for every small enough neighborhood of $p(x_1)$ is not homeomorphic to $(0,1)$ as if we remove a point we get more then 2 connected components.

  • 1
    $\begingroup$ Could you please explain more precisely why $p\mid_{(a,b)}$ must be injective? If not, you have distinct $x_1, x_2 \in (a,b)$ such that $p(x_1) = p(x_2) = y$, but why is it impossible that $y$ has a neighborhood homeomorphic to $(0,1)$? $\endgroup$
    – Paul Frost
    Aug 18, 2019 at 16:53
  • $\begingroup$ @PaulFrost I edited the answer to explain this $\endgroup$
    – Elad
    Aug 19, 2019 at 7:35
  • $\begingroup$ I do not think you can argue that there is a first (= smallest) element $x_2$, but that should be irrelevant. Since $x_1,x_2 \in p^{-1}(U) = U \times F$, you find two intervals $J_k = (x_k - \epsilon,x_k+\epsilon) \subset U \times \{f_k\}$ which are mapped by $p$ homeomorphically onto $U_k = p(J_k)$. But then obviously $p$ is not injective on $(x_1-\varepsilon,x_1)\cup (x_1, x_1 +\varepsilon)\cup(x_2 -\varepsilon,x_2)$. $\endgroup$
    – Paul Frost
    Aug 19, 2019 at 8:38

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