fundamental group of suspension about book

[mikeshulman]

Yeah. What is the right thing to say here? We have "words" in elements of
\pi0(A) and their inverses, but they are all required to start with an
element of \pi0(A) and end with the inverse of such. Maybe if we fix an
element a:\pi0(A), we can say that it is the free group on \pi0(A)
quotiented by the relation a=1? Which should be the free group on the
complement of a, if such exists?

On Mon, Mar 25, 2013 at 4:25 PM, Daniel R. Grayson <[email protected]

wrote:

The sentence

By comparing universal properties, we can identify this with the free
group on the set $\pi_0(A)$.

in homotopy.tex is not quite right. For example, if A is S^0, we would get
a free group with 2 generators, which is not Z, the fundamental group of
S^1.

(The context is taking the pushout 1 <-- A --> 1 and computing its
fundamental group.)

—
Reply to this email directly or view it on GitHubhttps://github.com//issues/24
.

[guillaumebrunerie]

If A is pointed, then it should be the free group on the pointed set A,
where in particular the base point of A is required to be the identity of
the group.

Guillaume
On Mar 25, 2013 5:04 PM, "Mike Shulman" [email protected] wrote:

Yeah. What is the right thing to say here? We have "words" in elements of
\pi0(A) and their inverses, but they are all required to start with an
element of \pi0(A) and end with the inverse of such. Maybe if we fix an
element a:\pi0(A), we can say that it is the free group on \pi0(A)
quotiented by the relation a=1? Which should be the free group on the
complement of a, if such exists?

On Mon, Mar 25, 2013 at 4:25 PM, Daniel R. Grayson <
[email protected]

wrote:

The sentence

By comparing universal properties, we can identify this with the free
group on the set $\pi_0(A)$.

in homotopy.tex is not quite right. For example, if A is S^0, we would
get
a free group with 2 generators, which is not Z, the fundamental group of
S^1.

(The context is taking the pushout 1 <-- A --> 1 and computing its
fundamental group.)

—
Reply to this email directly or view it on GitHub<
https://github.com/HoTT/book/issues/24>
.

—
Reply to this email directly or view it on GitHubhttps://github.com//issues/24#issuecomment-15425451
.

[mikeshulman]

Is that the same as what I said?

[guillaumebrunerie]

If the complement of the base point exists, then the pointed free group is
a point plus the unpointed free group of the complement.

I'm not sure what you mean by adding the relation a=1, if you do that after
"free grouping", then what you get is not a group anymore. If you do it at
the same time (defining the free group as a higher inductive type) it
should be the same.
On Mar 25, 2013 5:16 PM, "Mike Shulman" [email protected] wrote:

Is that the same as what I said?

—
Reply to this email directly or view it on GitHubhttps://github.com//issues/24#issuecomment-15426142
.

[mikeshulman]

Of course you have to do it at the same time. I meant "relation" in the sense of presenting a group by generators and relations.

[mikeshulman]

I think one good answer to this and also the other issue about free products is to discuss colimits and presentations of groups as an example around abouts section 5.10. Then we will have all the constructions of these things in hand already, by the time we need them in this chapter.

[DanGrayson]

The words alternate between elements of pi_0 and their inverses, too. So
it's the free group on pairs (a,b) with a,b in pi_0 and with (b,a)
identified with the inverse of (a,b) and with (a,a) identified with 1 and
with (a,b)(b,c) identified with (a,c). (I'm trying to avoid picking a base
point for A.)

On Mon, Mar 25, 2013 at 5:04 PM, Mike Shulman [email protected]:

Yeah. What is the right thing to say here? We have "words" in elements of
\pi0(A) and their inverses, but they are all required to start with an
element of \pi0(A) and end with the inverse of such. Maybe if we fix an
element a:\pi0(A), we can say that it is the free group on \pi0(A)
quotiented by the relation a=1? Which should be the free group on the
complement of a, if such exists?

On Mon, Mar 25, 2013 at 4:25 PM, Daniel R. Grayson <
[email protected]

wrote:

The sentence

By comparing universal properties, we can identify this with the free
group on the set $\pi_0(A)$.

in homotopy.tex is not quite right. For example, if A is S^0, we would
get
a free group with 2 generators, which is not Z, the fundamental group of
S^1.

(The context is taking the pushout 1 <-- A --> 1 and computing its
fundamental group.)

—
Reply to this email directly or view it on GitHub<
https://github.com/HoTT/book/issues/24>
.

—
Reply to this email directly or view it on GitHubhttps://github.com//issues/24#issuecomment-15425451
.

[mikeshulman]

Have a look at the changes in d7eb4bb, and reopen if you aren't satisfied.