Equivalence of Covering Maps and Quotient Maps

[nigelscott]

I am newbie to topology and trying to understand covering maps and quotient maps. At first sight it seems the two are closely related. For example SO(3) is double covered by SU(2) and is also the quotient SU(2)/ℤ₂ so the 2 maps appear to be equivalent. Likewise, for ℝ and S¹. However, I don't think this is true in the general case. Can somebody provide an intuitive understanding of the criteria for equivalence and how it applies to one of the above examples.

 

[andrewkirk]

There is a big overlap between covering and quotient maps. Both are continuous and surjective.
But a quotient map has the property that a subset of the range (co-domain) must be open if its pre-image is open, whereas a covering map need not have that property,
and a covering map has the local homeomorphism property, which a quotient map need not have.

I think if either of them is injective then it will be a homeomorphic endomorphism of the space, and will be both a quotient and a covering map. But when it is not injective, it can be one and not the other.

An example of a quotient map that is not a covering map is the quotient map from the closed disc to the sphere $S^2$ that maps every point on the circumference of the disc to a single point P on the sphere. It can also be thought of as gluing together (identifying) all points on the disc's circumference. For that map, any sufficiently small open neighbourhood N of P will have a pre-image that is not homeomorphic to N because the pre-image will have a hole in it whereas N does not.

I can't think of an example of a covering map that is not a quotient map. Perhaps someone else can. Or perhaps it is the case that all covering maps are also quotient maps.

 

[Infrared]

There is a big overlap between covering and quotient maps. Both are continuous and surjective.
But a quotient map is open, whereas a covering map need not be

It's the other way around. Covering maps are open since they are local homeomorphisms. On the other hand, if $\sim$ identifies all the rational numbers of $\mathbb{R}$, then $\mathbb{R}\to\mathbb{R}/\sim$ is not open. It is true that quotients by topological group actions are open however.

I can't think of an example of a covering map that is not a quotient map. Perhaps someone else can. Or perhaps it is the case that all covering maps are also quotient maps.

Covering maps are always quotient maps. Let $p:X\to B$ be a covering, and $U\subset B$ is such that $f^{-1}(U)$ is open. But covering maps are open, so $f(f^{-1}(U))=U$ is open. Hence, $U\subset B$ is open iff $f^{-1}(U)$ is; that is, $B$ has the quotient topology. (This argument works for any continuous map that is open and surjective.)

 

[Infrared]

One common way in way covering maps come up is from group actions. If a finite group acts freely on a Hausdorff space by homeomorphisms, then the quotient map is a covering map. In your case the nontrivial element of $\mathbb{Z}_2$ acts $SU(2)$ by changing signs. The double cover $SU(2)\to SO(3)$ is actually a group homomorphism with kernel $\{\pm I\}$, meaning that two elements in the same orbit by $\mathbb{Z}_2$-action have the same image in $SO(3)$, so we get our identification $SU(2)/\mathbb{Z}_2=SO(3)$.

 

[nigelscott]

OK. So covering maps are always quotient maps but quotient maps are not always covering maps. In the latter case I have read that q:X -> X/G is a covering map if and only if G acts properly discontinuously on X. Properly discontinuously means that for every x∈X there is a neighborhood U such that gU∩U=∅ for all g∈G−{e}. What exactly does this last sentence mean. Also, I am a little confused about the difference between open sets and open neighborhoods in this context.

 

[Infrared]

A neighborhood of a point $p$ is any set that contains an open set containing $p$. This means that an open neighborhood of a point is just an open set which contains it.

Which part of the definition of a properly discontinuous action do you have trouble with?

You should also check my criterion of a finite group acting freely on a Hausdorff space is an example of a properly discontinuous action. It is an important special case.

 

[nigelscott]

I interpret (gU)∩U = 0 to mean all non trivial elements of G move U outside of itself. Can I interpret this as being equivalent to the disjoint union of sets in the covering space?

I think if either of them is injective then it will be a homeomorphic endomorphism of the space, and will be both a quotient and a covering map. But when it is not injective, it can be one and not the other.

I am confused about this statement.

 

[Infrared]

I interpret (gU)∩U = 0 to mean all non trivial elements of G move U outside of itself.

This is correct (except that you meant to write $\emptyset$ instead of 0).

Can I interpret this as being equivalent to the disjoint union of sets in the covering space?

Yes, though this isn't necessary for the definition. Let the quotient map be $p:X\to X/G$. If $z\in X/G$ and $p(x)=z$, then we can form a neighborhood $U$ around $x$ satisfying this property. Then $V=p(U)$ is an open neighborhood of $z$ such that $p^{-1}(V)=\bigcup_{g\in G} gU$. This union is disjoint by the assumption on $U$, and each $gU$ is taken to $V$ homeomorphically by $p$.

I am confused about this statement.

If a quotient map is injective, then it is a homeomorphism. If a covering map is injective, then it is an homeomorphism. You should verify both of these statements.

 

[nigelscott]

Thanks for your help. This is much clearer now.

 

[lavinia]

A surjective map $π:X→Y$ is a quotient mapping if a set $U$ in $Y$ is open if and only if its inverse image $π^{-1}(U)$ is open in $X$.

This is clearly true of a covering projection but a quotient mapping need not be a covering projection.

Here are some examples of quotients that are not coverings.

Map $X$ onto a single point. This is not a covering except if $X$ has the discrete topology.

If instead of $SU(2)/Z_2$ one takes $SU(2)/SO(2)$. More generally any Lie group modulo a non-discrete subgroup is a quotient space but the projection map is not a covering.

Common important quotients are gluing maps. For instance a cylinder is the quotient of a square with two opposite edges glued together.

A Mobius band is the quotient of a square with two opposite edges glued - but with a half twist.

A sphere is the quotient of two disks with their boundary circles identified.

A sphere is topologically a tetrahedron which is the quotient of four triangles with appropriate edge identifications.


Other common examples are:

A disk is the quotient of the Cartesian product $S^1×[0,1]$ by identifying the circle $S^1×1$ to a point.

More generally if the circle is replaced by an arbitrary topological space then one gets the "cone" on $X$.
So the disk is the cone on a circle.

If one also identifies $X×0$ to a point one gets the "suspension" of $X$. The suspension of a circle is a 2 dimensional sphere. The suspension of an $n$ dimensional sphere is an $n+1$ dimensional sphere.