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Suppose I have two fibre sequences $F\to L\to K$ and $F'\to L'\to K'$ fitting into the commutative diagram: $\require{AMScd}$ \begin{CD} F' @>>> L' @>>> K'\\ @VV{\cong}V @VVV @VV{\cong}V\\ F @>>> L @>>> K \end{CD} Is it automatically true that the middle vertical map is also an isomorphism? This looks a bit like the five-lemma in homological algebra, does it hold in this "non-linear" context too?

If this works, I expect it to be well-known. Of course a reference for the result would be enough.

Remark: I am not interested in a homotopical version, I really want an isomorphism in the category of simplicial sets. If it can help, the middle map can be assumed to be injective.

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    Did you try chasing? Just because you're not in the perfect setting doesn't mean it won't work. – 2017-02-13
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    @PedroTamaroff I tried a bit, but it didn't work. Say I want to prove surjectivity. I can start with $x\in L_n$ and try to lift it to an element of $L'_n$, however once I've obtained an element an element $y'\in K'$ such that the images of $x$ and $y'$ in $K$ agree, I cannot lift it to an element of $L'$ unless $x$ actually comes from $F$. I'll think about it some more, but right now I don't know how to make it work. – 2017-02-13
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    Right, you're missing "exactness". – 2017-02-14
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    @PedroTamaroff Exactly. I have some kind of "exactness at the basepoint" but it is not enough because we are not in a linear setting. – 2017-02-14

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