Intuition behind induced homomorphism from covering maps

[PsychonautQQ]

Let $p: X-->Y$ be a cover map. Then the induced homomorphism will inject the fundamental group of X into the fundamental group of Y; furthermore, the image of the fundamental group of X under p will be a subgroup of the fundamental group of Y

I read the proofs, but I'd like to have an informal discussion of this matter to develop my intuition on the matter. First of all, it just seems weird that the fundamental group of X gets injected into Y, because X is a cover space of Y! X is so much bigger than Y! it covers it multiple times! Why should we be able to inject the fundamental group of X into the fundamental group of Y?

Yeah, If someone could help me develop my understanding of what's going on here that'd be awesome.

 

[andrewkirk]

It's a good question: why is the fundamental group of a big (covering) set C smaller than that of a related smaller (base) set B?

A series of observations might help.

The first is that a fundamental group pertains not to a set C but to a set-and-point combination $(C,c_0)$, with the point $c_0$ being the end point of the loops. By forcing the loops to start at the specified point we are in a sense defusing some of the size advantage of C because our choice of a start point is just as restricted in C as it is in B.

The second is that, given a point $b_0\in B$ and another point $c_0\in \phi^{-1}(b_0)$ where $\phi:C\to B$ is the covering map, there is a natural bijection between the set $P_{c_0}$ of paths in C starting at $c_0$ and the set $P_{b_0}$ of paths in B starting at $b_0$. So there are in a sense the same number of paths of interest in the two spaces.

Thirdly we notice that the set $L_{b_0}$ of loops in B based at $b_0$ is a subset of the set of paths $P_{b_0}$ but it is not in bijection with the set $L_{c_0}$ of loops in C at $c_0$, because most of the inverse images under $\phi$ of loops in $L_{b_0}$ are paths but not loops in C. In the classic case of $\mathbb R$ covering $S^1$, a loop is characterised by the net number of times it goes around the circle, and in which direction. But the pre-image of those loops under $\phi$ is only a loop in the covering set $\mathbb R$ when the net number of windings is zero. If the net number of windings is nonzero, the corresponding path in C will not be a loop.
So there are in a sense more loops in B at $b_0$ that there are in C at $c_0$.

A fundamental group is just a collection of equivalence classes of loops so, given $(B,b_0)$ has more loops than $(C,c_0)$, we would intuitively expect it to have more equivalence classes - i.e. a bigger fundamental group.

 

[lavinia]

Intuitively a covering is a wrapping of one space around another so some line segments that may not be loops can become loops.

It is perhaps less intuitive that if the projection of a loop to the base is null homotopic then it is null homotopic to begin with. This follows from a general theorem called the Covering Homotopy Theorem. The proof is technical and the idea depends on the local triviality of the covering projection. Suppose a loop in the base is null homotopic and the homotopy lies completely inside an open set $U$ whose inverse image in the total space is a disjoint union of open sets $V_{i}$ that are each homeomorphic to $U$.Then the homotopy obviously lifts to anyone of the $V_{i}$. If the homotopy does not lie completely in $U$ then some part of it does and this part can be lifted. Completely cover the image of the homotopy with more open sets like $U$ and the part of the homotopy that lies in anyone of them can be lifted. Stitch the pieces together to lift the entire homotopy.

-Examples:

-The sphere is a two fold cover of the projective plane. The sphere is simply connected. The projective plane has fundamental group $Z_2$. An arc that connects two opposite poles on the sphere becomes a loop in the projective plane because the two poles are projected to the same point.

- The plane is an infinite cover of the torus. The plane is simply connected. The fundamental group of the torus is $Z×Z$. If one makes a torus by identifying all points in the plane whose xy- coordinates both differ by integers then all horizontal and vertical lines become closed curves on the torus. Additionally, any straight line that makes a rational angle with the x-axis becomes a closed curve on the torus.

The torus is a two fold cover of the Klein bottle. The fundamental group of the torus is $Z×Z$. The fundamental group of the Klein bottle is an extension of $Z×Z$ by $Z_2$. That is: there is a short exact sequence of groups $0→Z×Z→K→Z_2→0$. $K$ is not abelian. The first homomorphism, $Z×Z→K$ ,is the induced map on fundamental groups.

The torus is also a two fold cover of a torus. Again one has a short exact sequence. It is $0→Z×Z→Z×Z→Z_2→0$.

 

[dkotschessaa]

I like this question and the replies so far. I'm wondering, does it help if we think of the trivial case where a space is a covering of itself? What's generally the first non-trivial covering space we look for? Are n copies of a space X a covering space for X? (I don't think this is true)

 

[fresh_42]

I tried to imagine a twisted circle in space which projects to an eight in the plane and @lavinia's description in the first paragraph reminded me on a stack of record discs used to play 10 or so in a row without the need of manual changes.

 

[lavinia]

I like this question and the replies so far. I'm wondering, does it help if we think of the trivial case where a space is a covering of itself? What's generally the first non-trivial covering space we look for? Are n copies of a space X a covering space for X? (I don't think this is true)

A circle is a covering of itself - as many time as you want. This is not a trivial covering.

 

[PsychonautQQ]

It's a good question: why is the fundamental group of a big (covering) set C smaller than that of a related smaller (base) set B?

The second is that, given a point $b_0\in B$ and another point $c_0\in \phi^{-1}(b_0)$ where $\phi:C\to B$ is the covering map, there is a natural bijection between the set $P_{c_0}$ of paths in C starting at $c_0$ and the set $P_{b_0}$ of paths in B starting at $b_0$. So there are in a sense the same number of paths of interest in the two spaces.

.

The natural bijection between paths in the two spaces... This is obvious because every path in $B$ starting at $b_{0}$ has a unique lift to a path in $C$ starting at $c_{0}$? Or does that unique lift only apply to loops in $B$? I guess I'm a little confused at this natural bijection, it seems like there should be more paths in $C$ than in $B$ because, well, $C$ is bigger. That being said, I guess that doesn't matter since a path can go back and forth and what not. What makes this bijection so natural exactly?

 

[PsychonautQQ]

It's a good question: why is the fundamental group of a big (covering) set C smaller than that of a related smaller (base) set B?Thirdly we notice that the set $L_{b_0}$ of loops in B based at $b_0$ is a subset of the set of paths $P_{b_0}$ but it is not in bijection with the set $L_{c_0}$ of loops in C at $c_0$, because most of the inverse images under $\phi$ of loops in $L_{b_0}$ are paths but not loops in C. In the classic case of $\mathbb R$ covering $S^1$, a loop is characterised by the net number of times it goes around the circle, and in which direction. But the pre-image of those loops under $\phi$ is only a loop in the covering set $\mathbb R$ when the net number of windings is zero. If the net number of windings is nonzero, the corresponding path in C will not be a loop.
So there are in a sense more loops in B at $b_0$ that there are in C at $c_0$.

Right, this makes so much sense actually. The Given the preimage of a loop in $B$, a lot of those will be paths in $C$ and not loops... awesome thanks!

 

[dkotschessaa]

A circle is a covering of itself - as many time as you want. This is not a trivial covering.

Ok, that's what I was asking. I thought it was either a) not true at all or b) so obvious that nobody bothered to say it that way.

So we have the trivial covering p:X-->X and then the 'first nontrivial coverings" (n copies of X) ---> X

I wonder if this can help our intuition a bit.

 

[PsychonautQQ]

Intuitively a covering is a wrapping of one space around another so some line segments that may not be loops can become loops.

It is perhaps less intuitive that if the projection of a loop to the base is null homotopic then it is null homotopic to begin with. This follows from a general theorem called the Covering Homotopy Theorem. The proof is technical and the idea depends on the local triviality of the covering projection. Suppose a loop in the base is null homotopic and the homotopy lies completely inside an open set $U$ whose inverse image in the total space is a disjoint union of open sets $V_{i}$ that are each homeomorphic to $U$.Then the homotopy obviously lifts to anyone of the $V_{i}$. If the homotopy does not lie completely in $U$ then some part of it does and this part can be lifted. Completely cover the image of the homotopy with more open sets like $U$ and the part of the homotopy that lies in anyone of them can be lifted. Stitch the pieces together to lift the entire homotopy.

-Examples:

-The sphere is a two fold cover of the projective plane. The sphere is simply connected. The projective plane has fundamental group $Z_2$. An arc that connects two opposite poles on the sphere becomes a loop in the projective plane because the two poles are projected to the same point.

- The plane is an infinite cover of the torus. The plane is simply connected. The fundamental group of the torus is $Z×Z$. If one makes a torus by identifying all points in the plane whose xy- coordinates both differ by integers then all horizontal and vertical lines become closed curves on the torus. Additionally, any straight line that makes a rational angle with the x-axis becomes a closed curve on the torus.

The torus is a two fold cover of the Klein bottle. The fundamental group of the torus is $Z×Z$. The fundamental group of the Klein bottle is an extension of $Z×Z$ by $Z_2$. That is: there is a short exact sequence of groups $0→Z×Z→K→Z_2→0$. $K$ is not abelian. The first homomorphism, $Z×Z→K$ ,is the induced map on fundamental groups.

The torus is also a two fold cover of a torus. Again one has a short exact sequence. It is $0→Z×Z→Z×Z→Z_2→0$.

Man i love this website!

 

[lavinia]

I tried to imagine a twisted circle in space which projects to an eight in the plane and @lavinia's description in the first paragraph reminded me on a stack of record discs used to play 10 or so in a row without the need of manual changes.

https://math.stackexchange.com/questions/354056/universal-cover-of-a-figure-eight

 

[lavinia]

Ok, that's what I was asking. I thought it was either a) not true at all or b) so obvious that nobody bothered to say it that way.

So we have the trivial covering p:X-->X and then the 'first nontrivial coverings" (n copies of X) ---> X

I wonder if this can help our intuition a bit.

You might find it helpful to look at fundamental domains of coverings. A fundamental domain is a set in the covering space that maps bijectively onto the base space. Usually a fundamental domain can be chosen to be path connected so it is often easy to visualize. The base space is equal to the fundamental domain (if chosen carefully) modulo identifications on its boundary. One can get an idea of the covering from these identifications.

For instance, one fundamental domain for the cover $x→e^{2πix}$ of the circle by the line is the half open interval $[0,1)$. The circle is obtained by identifying the boundary points $0$ and $1$.

The covering $(x,y)→(e^{2πix},e^{2πiy})$ of the torus by the plane has fundamental domain a half open square with sides of length 1 and edges touching the x and y axes.

A fundamental domain for the covering $x→-x$ of the Projective Plane by the sphere is the open northern hemisphere plus half of the equator.

 

[fresh_42]

Another example are the (special) orthogonal groups with spin groups or the $SL_2(\mathbb{R})$.

 

[lavinia]

@dkotschessaa Here is a three dimensional covering space. The base space is a closed three dimensional manifold. While it can not be visualized (no closed 3 manifold can), the identifications show you how to construct it from a rectangular parallelepiped.

The total space is $R^3$. The group of covering transformations is the group generated by standard lattice of integer coordinate points viewed as a group of translations together with the map $σ:(x,y,z)→(x+1/4,-z,y)$. A fundamental domain is 1/4 of the unit cube sliced parallel to the yz plane.

 

[andrewkirk]

The natural bijection between paths in the two spaces... This is obvious because every path in $B$ starting at $b_{0}$ has a unique lift to a path in $C$ starting at $c_{0}$? Or does that unique lift only apply to loops in $B$?... What makes this bijection so natural exactly?

The unique lift applies to all paths starting at $b_0$, not just to loops. It is 'natural' in the sense that it doesn't require making any choices or additional definitions. Everything is set by the covering map $\phi$ and the choice of the point $c_0\in\phi^{-1}(b_0)$.

I guess I'm a little confused at this natural bijection, it seems like there should be more paths in $C$ than in $B$ because, well, $C$ is bigger.

This is where the first point in post 2 comes in. There are many paths in
C that are pre-images under $\phi$ of $\gamma$. But fixing the point $c_0$ eliminates all but one of them. Only one of those pre-image paths starts at that point. The others all start at different points in $\phi^{-1}(b_0)$.

 

[WWGD]

There is a recipe for constructing covering spaces of any space that is, I think just semilocally path-connected. You assemble the space from the paths lifted.