1
$\begingroup$

At some places like wikipedia i have read that the fundamental group of a topological group roughly count the holes in the topological space,but the fundamental group of a torus with one hole is free group on two symbols.so i am quite confused in understanding the above remark.

what is the justification of the remark that 'fundamental group roughly count the holes in the topological space' ? I don't understand this basically because any group can be the fundamental group of a space.For instance,suppose the fundamental group is lets say is $\frac {\mathbb Z}{n\mathbb Z}$ then what can we tell about the holes in the space?

Could someone explain me the meaning of above remark?

$\endgroup$
4
  • $\begingroup$ The torus has two holes, the doughnut hole, but also the doughnut itself is hollow this is the second hole. If the doughnut were solid its fundamental group would be $\mathbb{z}$. $\endgroup$ – Rene Schipperus Oct 7 '16 at 13:16
  • $\begingroup$ @ReneSchipperus: Does this thinking always works that fundamental group counts the holes in space? $\endgroup$ – Dontknowanything Oct 7 '16 at 13:19
  • $\begingroup$ @ Like all intuition is a bit rough, but this one is very good, however you might want to look into "the topologist sine curve", and decide for yourself. $\endgroup$ – Rene Schipperus Oct 7 '16 at 13:22
  • $\begingroup$ @ReneSchipperus: thanks,i'll try to look in that case. $\endgroup$ – Dontknowanything Oct 7 '16 at 13:28
1
$\begingroup$

You need to ask yourself first, what is the fundamental group? It is created from the collection of all paths $f:[0,1]\to X$ such that $f(0)=f(1)=p$ where $p$ is some point in $X$. But this is not enough as we say that two loops, as they are called, are equivalent if we can find another continuous function $F:[0,1]\times[0,1]\to X$ such that $F(0,x)=f(x)$ and $F(1,x)=g(x)$ where $f$ and $g$ are loops with the same base point.

To say it in different words, we can deform one loop into the other continuously which is the key in all of topology. If we restrict ourselves to a 2D surface for imagination purposes a loop that goes through a connected space with no holes in it, can be deformed into a single point. However if the loop makes a turn around a hole before returning to the base point, then we cannot deform this loop to get away from being around the hole, without breaking it apart somehow, so such a transformation is not allowed.

Now another hole in the surface will give the loop another oppertunity to pass by it and in turn preventing a continuous deformation to get rid of that passing of the hole.

In that way the fundamental group "counts the holes" because each of the supposed holes prevents the loop to be retracted to the basepoint in one manner and as such the group becomes non-trivial.

$\endgroup$
11
  • $\begingroup$ Thank you..suppose the fundamental group is lets say is $\frac {\mathbb Z}{3 \mathbb Z}$ then what can we tell about the holes in the space? $\endgroup$ – Dontknowanything Oct 7 '16 at 14:14
  • $\begingroup$ You don't get that to my knowledge, they are usually in the form of $\Bbb Z^n$ where $n$ would be the number of distinct holes in our euclidean plane example $\endgroup$ – Zelos Malum Oct 7 '16 at 14:17
  • 2
    $\begingroup$ Keep in mind that there is a reason they said "Roughly" $\endgroup$ – Zelos Malum Oct 7 '16 at 14:23
  • 1
    $\begingroup$ You need to remember that the concept of "holes" isn't necciserily applicable then, the idea of holes comes from the euclidean space where we cut out a slice and such. That is why one says "roughly" because in the nicer types of topological spaces there is a correspondance but less nice and trivial ones, the equivalence gets muddy and even lost. $\endgroup$ – Zelos Malum Oct 7 '16 at 14:26
  • 1
    $\begingroup$ One can interpret a fundamental group of $\mathbb{Z}/n\mathbb{Z}$ as meaning that one has a fractional hole, in the sense that a loop going around the hole really is a hole, but if you go round it $n$ times, then actually you're no longer going around the hole at all as the loop is homotopic to the identity. (this is of course only a vague 'intuitionist' interpretation, and not rigorous). $\endgroup$ – Dan Rust Oct 7 '16 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.