Today, we will define *relative* versions of the homotopy groups, and show that they fit into an exact sequence. So let be a pointed space and a subspace containing the basepoint. Let . Then we define

to be the pointed homotopy class of maps . Here has a basepoint, which is located on the boundary .

Definition 1is called the -threlative homotopy groupof the pair . (We have not yet shown that it is a group.)

Another perhaps more geometric way of thinking of the relative homotopy groups is as follows. Namely, it is the homotopy class of maps , where is the -cube and is the complement of the front face . So on the boundary, such a map is except possibly on the front face, where it is at least in . The reason is that if we quotient by , we get the pair .

And this is also how we can make into a group if . Namely, given two maps , we join the two cubes together among a face that is not the front face or the back. This is possible if . Here is a picture.

We can also do this more formally. One way is to note that if we define to be the family of paths that start in and end in , topologized with the compact-open topology and basepointed with the constant path at , then is just . The addition can be defined in this way, by the additive structure on . In particular, we have seen that is a group when . It is even abelian when .

The main result we want to prove today is:

Theorem 2Let be a closed subspace. Then there is an exact sequence of homotopy groups:

Before proving it, let’s at least describe what the boundary maps look like. (The maps come by functoriality, and the maps are obvious.) So say we have a class in represented by . Then takes values in and is on the boundary on . This is what we define to be the boundary of the homotopy class. Moreover, this way of defining the map is easily seen to be a group-homomorphism when .

Using the exact sequence for a pair and some purely algebraic nonsense, one can show:

Theorem 3Let be a triple. Then there is a long exact sequence

This is a standard argument in axiomatic homology theory, where you go from the exact sequence of a pair to the exact sequence of a triple. (In *singular* homology theory, the exact sequence of a triple can of course be proved directly.) I won’t try to blog about the argument, because it’s really messy and completely un-topological.

** The abstract approach **

As in Hatcher, one can argue geometrically about the exact sequence. There is also the categorical approach. Since this is (to me, at least) harder to understand, I will blog about it.

The first thing to do is to understand the homotopy groups and relative homotopy groups in such a manner as to apply the Barratt-Puppe sequence. So start with the inclusion of pairs . The mapping cone (i.e., pairs of mapping cones) is , where is a proper (and consequently) subarc of . So we can replace the last term by for the basepoint.

But wait, you say. We never talked about *pairs* before, right? OK, perhaps we should explain what’s going on in outline.

Let us consider the category of *pairs* of pointed topological spaces and basepoint homotopy classes of maps of pairs. So two maps are homotopic if there exists such that , , and . This is the pair analog of the category . It too has an initial object (namely, ), so it makes sense to talk about coexact sequences in this category.

Next, we note that there is a version of the Barratt-Puppe sequence for pairs.

Theorem 4 (Barratt-Puppe)Let be a morphism. There is then a coexact sequence in

I can’t see how this theorem is a corollary of the theorem I stated last time. Actually, perhaps I should have been more thoughtful and discussed it the first time in the context of pairs instead of the context of one pointed space. But anyway, it turns out that the proof is just the same. So I will not write it out here.

This implies that, as above, we have a coexact sequence beginning . Let us continue this. It goes

But these elements are very nice. In fact, we know that the reduced suspension of an -sphere is just an sphere. Actually, this is not completely trivial because of the reduction, but I will omit the proof. (You can look it up in Bredon’s book or Switzer’s book, for instance.) The suspension of a point, by contrast, is just a point. Finally, what is ? It turns out to be, up to homotopy, . This is clear if is replaced with the “unreduced” suspension; it is a general fact that for reasonable (well-pointed, to be precise) spaces, the unreduced suspension is homotopy equivalent to the regular reduced suspension.

Now we know that this sequence is coexact. So let’s take a pair . When we hom out in from the long sequence into the pair , we get an exact sequence of pointed sets. The associated sequence of pointed sets is, however, just

This is why we have an exact sequence for a pair.

Next time, we’ll discuss a *compression *criterion for deciding when a map represents zero in the relative homotopy groups. After that, over the next several posts, I would like to talk about some of the homotopy theory of CW complexes: the Whitehead theorem, CW approximation, Eilenberg-Maclane spaces, homotopy excision, and the Brown representability theorem. I will also need to go back to general topology to discuss fibrations (the dual idea to cofibrations) and how this long exact sequence of homotopy groups turns into something special when you have a fibration.

October 11, 2010 at 8:34 am

[…] one will be a quick post. In effect, we continue with last time, where we defined the relative homotopy groups, and now describe a practical means of determining when something in one of these relative groups […]

May 14, 2012 at 1:50 pm

It is sensational, everything is very well explained.

Please, can you help me with two problem? My problems are:

PROBLEM 1

Un espacio X se dice n-simple ([tex]n\geq 1[/tex]) si para cada par de puntos [tex]x_1,x_2 \in X[/tex] y caminos [tex]\gamma, \gamma ‘[/tex] de [tex]x_1[/tex] a [tex]x_2[/tex] se tiene [tex]\varphi_{[\gamma]} = \varphi_{[\gamma ‘]}[/tex], donde [tex]\varphi_{[\gamma]}, \varphi_{[\gamma ‘]}: \pi_n(X,x_2) \to \pi_n(X,x_1)[/tex] son los correspondientes isomorfismos de cambio de punto de base. Probar:

(a) X es n-simple si y solo si existe [tex]x_0 \in X[/tex] tal que [tex]\varphi_{[\gamma]}=id_{\pi_n(X,x_0)}[/tex] para todo [tex][\gamma] \in \pi_1(X,x_0)[/tex].

([b]Ayuda:[/b] Dados [tex]x_0,x_1,x_2 \in X[/tex] y caminos [tex]\gamma, \gamma ‘[/tex] de [tex]x_1[/tex] a [tex]x_2[/tex], considerar [tex]\xi = \eta + (\gamma + \bar{\gamma ‘}) + \bar{\eta}[/tex], siendo [tex]\eta[/tex] un camino de [tex]x_0[/tex] a [tex]x_1[/tex], para concluir [tex]\varphi_{[\gamma]} = \varphi_{[\gamma ‘]}[/tex].

(b) Concluir que si X es simplemente conexo, entonces X es n-simple para todo [tex]n\geq 1[/tex].

(c) Si X es n-simple, existe una correspondencia biyectiva entre los conjuntos de clases de homotopía [tex][S^n,p_0;X,x_0] \to [X^n;X][/tex].

PROBLEM 2

Consideremos aplicaciones continuas [tex]f_n:(B^n,S^{n-1}) \to (S^n,p_0)[/tex] tales que [tex]f|_{int(B^n}:int(B^n) \to S^n-\{p_0\}[/tex] es un homeomorfismo. Entonces, [tex]f_nxf_m: B^n x B^m \to S^n x S^m[/tex] envía [tex]\partial(B^n x B^m)=B^n x S^{m-1} \cup S^{n-1} x B^m[/tex] en [tex]S^n \vee S^m = S^n x \{ p_0 \} \cup \{ p_0 \} x S^m[/tex]. Elijamos un homeomorfismo [tex]\varphi: B^{n+m} \to B^n x B^m[/tex]. Comprobar que [tex]S^n x S^m[/tex] es homeomorfo al espacio obtenido al pegar una (n+m)-celula a [tex]S^n \vee S^m[/tex] mediante la aplicaci[color=red]ó[/color]n [tex](f_n x f_m) \circ \varphi |_{S^{n+m-1}}[/tex]

Thanks a lot!

May 14, 2012 at 1:55 pm

Oh, i thought that it was LaTex, and i wrote so, but i see that no, i’m sorry because you can’t understand nothing with the code… If you write this in a editor of LaTex, you can see well. Sorry….

Thanks of anyway and i hope your answer!

May 14, 2012 at 8:37 pm

Unfortunately I don’t read Spanish; could you perhaps translate?

January 25, 2014 at 11:11 am

Hello Akhil, you say “here is a picture” but no picture is shown 🙂