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Let $f:X\to Y$ be a map between connected CW complexes and $k\geq 0$ an integer. I am confused by the definition of $k$-connectivity or more fundamentally by what induces a long exact sequence of homotopy groups.

My favourite definition for $k$-connectivity is this: $f$ is called $k$-connected if the homotopy fiber $F$ of $f$ is $k-1$-connected, meaning that $\pi_i(F)=0$ for all $i$ with $0\leq i\leq k-1$. Of course, this definition is only reasonable for connected spaces $X$ and $Y$.

I know that for $F\to X\to Y$, there is a long exact sequence \begin{equation} \ldots\to\pi_i(F)\to\pi_i(X)\to\pi_i(Y)\to\ldots\to \pi_0(X)\to\pi_0(Y) \end{equation} by arguments about the homotopy fiber $F$. I like to define the relative homotopy groups $\pi_i(Y,A)$ as $\pi_{i-1}(F)$ and one gets from the above long exact sequence a long exact sequence for the relative homotopy groups.

Now Wikipedia defines for an inclusion $f:X\hookrightarrow Y$ to be $k$-connected, if its homotopy cofiber $C$ (= mapping cone) is $n$-connected, meaning that $\pi_i(C)=0$ for all $i$ with $0\leq i\leq k$. Even worse for me, the same Wikipedia article asserts a long exact sequence \begin{equation}(*)\hspace{10ex} \pi_i(X)\to\pi_i(Y)\to \pi_i(C) \end{equation} (however this is prolonged to the left and to the right).

My main question is: How do the two definitions of $k$-connectivity relate?

Maybe however, my problem of understanding begins even earlier: How do $\pi_i(C)$ and $\pi_i(Y,X)$ (from de definition above) relate?

I was able to show that for the connectivity \begin{equation} conn(F)+1=conn(C) \end{equation} holds for simply connected $X$ and $Y$. This means, that for simply connected $X$ and $Y$, the two definitions of $k$-connectivity of $f$ coincide if there is really an exact sequence (*). But what happens when $X$ and $Y$ are not simply connected but only connected?

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There is no long exact sequence in homotopy for a cofibration, so I think something is wrong in the wikipedia definition. –  Juan S Aug 28 '12 at 23:01
Wikipedia doesn't seem to claim the sequence $(*)$ extends on both sides. –  Thomas Belulovich Aug 29 '12 at 11:08

3 Answers 3

up vote 5 down vote accepted

Wikipedia's definition is (at the time of writing) not correct, in the sense that it's not equivalent to the usual condition on the relative homotopy groups (or the homotopy groups of the fiber) unless you assume simple connectivity.

Here's an example. It's a standard example of a CW-inclusion $A \to X$ which is an isomorphism on $\pi_1$ and on homology, but which is not surjective on $\pi_2$. If you believe that such a map $f$ exists then you can skip the below construction, because the mapping cone $Cf$ is simply-connected and has trivial homology groups, so by the Hurewicz theorem and the Whitehead theorem it is contractible; however, the map $f$ is not $2$-connected.

Let $W = S^1 \vee S^2$, whose universal cover is a copy of $\mathbb{R}$ with a copy of $S^2$ attached at each integer. The fundamental group of $W$ is $\mathbb{Z}$, and I'll call the generator $t$; it acts on the universal cover by translation by $1$. The Hurewicz map $\pi_2(W) \to H_2(W) = \mathbb{Z}$ is a "collapse" obtained by setting $t=1$.

The second homotopy group $\pi_2(W)$ is isomorphic to the group $\mathbb{Z}[t^{\pm 1}]$ as a group acted on by the fundamental group. Let $\alpha:S^2 \to W$ be a map whose image in this group is $t-2$. We can then form a space $$ X = W \cup_{\alpha} e^3 $$ formed by gluing a 3-cell in with attaching map $\alpha$. There is an inclusion $S^1 \subset W \subset X$.

From this point, I need some calculations.

  • The first part of the calculation is that the map $S^1 \to X$ is an isomorphism on homology groups; this follows by calculating cellular homology, because the map $\alpha$ is an isomorphism on $H_2$.
  • The second part of the calculation is that the maps $S^1 \to W \to X$ are all isomorphisms on $\pi_1$.
  • The third part of the calculation is that the second homotopy group of $X$ is the cokernel of the map $$ x \mapsto (t-2)x: \mathbb{Z}[t^{\pm 1}] \to \mathbb{Z}[t^{\pm 1}]. $$ This group is isomorphic to $\mathbb{Z}[1/2]$. Probably the most direct way is to take the universal cover of $X$, which is formed by taking the universal cover of $W$ and gluing in a $\mathbb{Z}$ worth of copies of $D^3$ along the translates of $\alpha$.

As a result, the map $\pi_2 S^1 \to \pi_2 X$ is the map $0 \to \mathbb{Z}[1/2]$, and so the second relative homotopy group is nonzero.

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I see, thanks. When I assume simple connectivity of $X$ and $Y$, I nonetheless need a ''relative'' Hurewicz theorem or a homotopy excision theorem, correct? –  Daniel Dreiberg Sep 3 '12 at 18:44
@DanielDreiberg Yes, that's correct. There's a very general relative Hurewicz theorem, which says that for an n-connected map, the $(n+1)$'st relative homology group is the first one, and it's the quotient of the (abelianization of) the first relative homotopy group by the action of $\pi_1$. –  Tyler Lawson Sep 4 '12 at 1:02

I've checked a number of sources and I can't find any reference to the homotopy cofiber when talking about $n$-connectedness. Here is a nice summary

1) A space is $n$-connected if $\pi_k X = 0$ for $k \le n$

2) A pair $(X,A)$ is $n$-connected if $\pi_k(X,A)=0$ for $k \le n$

3)A map $f:X \to Y$ is $n$-connected if the pair $(M_f,X)$ is $n$-connected, where $M_f$ is the mapping cylinder.

One can show then, using the long exact sequence of a pair, that this implies that $\pi_k(A) \simeq \pi_k(X)$ for $k < n$ and $\pi_n(A) \to \pi_n(X)$ is surjective.

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Hopefully I haven't made any fencepost errors below:

By the Blakers-Massey theorem, if the map $X \to Y$ is $n-$connected and $X \to *$ is $m-$connected (alternately, $X$ is $(m-1)$-connected), then the pushout square formed by $* \gets X \to Y$ is $(m+n-1)$-cartesian, meaning the map from $X$ to the homotopy pullback of $* \to C \gets Y$ is $(m+n-1)$-connected.

If $F$ is the homotopy fiber of $X \to Y$ (assume Y is pointed and connected so that this makes sense), then the square $(F \to *) \to (X \to Y)$ is a homotopy pullback square and so is $\infty$-cartesian.

Sitting these squares atop each other, we see the square $(F \to *) \to (* \to C)$ is $(m+n-1)$-cartesian, and so the map $F \to \Omega C$ is $m+n-1$-connected.

Alternatively, the map $\pi_k(F) \to \pi_{k+1}(C)$ is going to be an isomorphism for $k \le m+n-2$, and a surjection for $k = m+n-1$.

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Thank you for your answer. Im my order of learning, all this connectivity issues are used for the Blakers-Massey theorem. Also, I cannot believe that such a big cannon is really necessary to understand such a basic definition as connectivity of a map. Is it possible to use the Hurewicz theorem instead? –  Daniel Dreiberg Aug 29 '12 at 6:08
@Daniel: We want to call a pair $(X,A)$ $n$-connected if $\pi_i(X,A) = 0$ for all $i \le n$. Given your definition of $\pi_i(X,A)=\pi_i(F)$ this corresponds to the homotopy fiber being $n-1$ connected. The part on Wikipedia about the homotopy cofiber is simply wrong in my opinion - there is no long exact sequence as claimed (e.g. $S^1 \to D^2 \to S^2$) –  Juan S Aug 29 '12 at 6:37

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