Consider the commutative diagram below with exact rows (from the long exact sequence of homotopy groups) and $f_1,f_2,f_4,f_5$ bijective ($f_1,f_2$ homomorphisms). Does it follow that $f_3$ is also bijective?

\begin{align} \matrix{ \pi_1(A)&\to&\pi_1(X)&\to&\pi_1(X,A)&\to&\pi_0(A)&\to&\pi_0(X) \cr \downarrow f_1&&\downarrow f_2&&\downarrow f_3&&\downarrow f_4&&\downarrow f_5 \cr\pi_1(B)&\to&\pi_1(Y)&\to&\pi_1(Y,B)&\to&\pi_0(B)&\to&\pi_0(Y)} \end{align}

The usual proof of the five-lemma by diagram chasing makes use of the fact that the consituents are groups and all maps involved are homomorphisms. Since there is no group structure for the six sets on the right ($\pi_0$ and relative $\pi_1$), it does not apply.

However, with arguments about connected components and loops/paths, it is easily shown in an analogous diagram-chasing fashion that $f_3$ has to be surjective. The tricky part seems to be the injectivity (if it works at all): In analogy to the usual five-lemma, one can show that $f_3[x]=0\Rightarrow[x]=0$, but of course this does not immediately give injectivity due to $f_3$ only being a map between sets.

Is there a way to show injectivity? If not, is there a counterexample where this adapted five-lemma fails?


  • 3
    $\begingroup$ Can't you use the action of $\pi_1(X)$ on $\pi_1(X,A)$ to prove injectivity? $\endgroup$ – Ulrich Pennig Oct 17 '11 at 15:01
  • $\begingroup$ The following is the only way I see how this action could enter: If $f_3[x_1]=f_3[x_2]$, then the endpoints of $[x_1]$ and $[x_2]$ need to be in the same connected component of $A$ due to commutativity and injectivity of $f_4$. Say $[x_1]\neq[x_2]$, then there is a class $[\gamma]\in\pi_1(X)$ such that its action on $\pi_1(X,A)$ gives $[\gamma]\cdot[x_1]=[x_2]$. I don't see how to continue from this point in order to show that $[x_1]=[x_2]$, or equivalently $[\gamma]=0$, and I assume the reason is the assumption about basepoints that seems to be required (see answer below). $\endgroup$ – Pierre Oct 18 '11 at 9:22

This is covered in Exercise 9 in Section 4.1 of Allen Hatcher's book (page 358). It turns out that if (for all choices of base-points) $f_1,f_2,f_4,$ and $f_5$ are all bijections then $f_3$ is a bijection. Note that you can also extend one more term to the right in your diagrams, so that $\pi_0(X) \rightarrow \pi_0(X,A) \rightarrow 0$ if you define $\pi_0(X,A,x_0) = \pi_0(X,x_0)/\pi_0(A,x_0)$

  • $\begingroup$ Thanks for the reference! I was hoping to get around the assumption about the basepoints, but it looks like this is not possible. $\endgroup$ – Pierre Oct 20 '11 at 12:04

The abstract case is handled in

R. Brown, ``Fibrations of groupoids'', J. Algebra 15 (1970) 103-132.

where a 5-lemma type result is Theorem 4.9. I find it easier to handle the abstract case rather than the topological example. You get injectivity on Ker( $\pi_1(X, A) \to \pi_0 A)$ and an Example 4.10 (for groupoids) shows you do not always get injectivity.

There are other applications of the exact sequences of a fibration of groupoids, see

Heath, P.R. and Kamps, K.H. "On exact orbit sequences", Illinois J. Math. 26~(1) (1982) 149--154.

One aim of all editions (from 1968) of the book Topology and Groupoids is to work in the algebraic context of groupoids where appropriate, to see which results are really on groupoids.

  • $\begingroup$ You can find the family of exact sequences of a fibration of groupoids also in my book "Topology and groupoids" which also includes a Mayer-Vietoris type sequence for a pullback of a covering morphism of groupoids. $\endgroup$ – Ronnie Brown Feb 19 '12 at 15:07
  • $\begingroup$ How does this give me injectivity of $f_3$? Injectivity on Ker $\pi_0A\to\pi_0X$ only makes sense for $f_4$, which is assumed to be injective on all of $\pi_0A$ anyway. It seems to me that you are referring to a situation with a shift to the right, where there is another $\pi_0$, in which the question is about bijectivity of $f_4$ given bijectivity of the surrounding maps - the same goes for the paper you cite. Or is there a way to apply these result here? $\endgroup$ – Pierre May 24 '12 at 15:49
  • $\begingroup$ @rkdy Sorry, and you are quite right. I have edited my answer to give the correct placing! What I suggest is that the abstract case gives an easy place to test hypotheses and proofs. Hope that helps. $\endgroup$ – Ronnie Brown May 28 '12 at 19:46

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