Five-lemma for the end of long exact sequences of homotopy groups 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?
Thanks!
 A: 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.  
A: 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 $\pi_0(X,A,x_0) = \pi_0(X,x_0)/\pi_0(A,x_0)$
