Timeline for Fundamental group of a topological pullback
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 31, 2013 at 8:30 | vote | accept | Mark Grant | ||
| May 31, 2013 at 1:17 | comment | added | Jeff Strom | This sequence is best (from my point of view) derived from the "Mayer-Vietoris fiber sequence" $\Omega Z \to E\to X\cross Y$ associated to the (homotopy) pullback square. This is in my book. | |
| May 30, 2013 at 18:14 | comment | added | Allen Hatcher | I agree that this should be in algebraic topology textbooks. The underlying algebra at least can be found in my book as Exercise 38 for section 2.2, although there is an implicit assumption there that all the groups are abelian. With just a little care the proof (diagram chasing) also works in the nonabelian case, with the caveat mentioned in earlier comments that the map from the product may not be a homomorphism. The exercise says in essence that the two long exact sequences of homotopy groups for the given fibration and its pullback combine to give Oscar's exact sequence. | |
| May 30, 2013 at 15:20 | comment | added | Oscar Randal-Williams | @Mark: I don't know a reference. I hadn't noticed, but the map $\pi_1(X,x) \times \pi_1(Y, y) \to \pi_1(Z, z)$ is not a homomorphism (it is given by the formula you wrote), and exactness at this point is in the sense of pointed sets. Note that the preimage of the identity under that map is the same as the pullback of $\pi_1(X,x) \to \pi_1(Z,z) \rightarrow \pi_1(Y,y)$, so is a group. | |
| May 30, 2013 at 15:00 | comment | added | Mark Grant | Also, do you know a reference for this sequence? | |
| May 30, 2013 at 14:59 | comment | added | Mark Grant | This is great, thanks. So are the maps in this sequence what you'd expect? In fact, I can see an obvious candidate for the map $\pi_1(E)\to \pi_1(X)\times \pi_1(Y)$, but not so much for the map $\pi_1(X)\times\pi_1(Y)\to \pi_1(Z)$. I want to send $(a,b)$ to $f_\sharp(a)\cdot g_\sharp(b)^{-1}$, but this won't be a homomorphism if $\pi_1(Z)$ isn't commutative. | |
| May 30, 2013 at 14:11 | comment | added | K.J. Moi | That sequence is quite useful, for some reason it usually doesn't appear in algebraic topology books. | |
| May 30, 2013 at 14:03 | history | answered | Oscar Randal-Williams | CC BY-SA 3.0 |