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This should be such an elementary problem in algebraic topology that I'm almost too embarrassed to ask, but here goes.

Let $f: X\to Z$ be a surjective fibration, and let $g: Y\to Z$ be any map. Assume all spaces are path-connected, and base points $x,y,z$ chosen so that $f(x)=g(y)=z$. Form the pullback in the topological category, $$ \begin{array}{ccc} E & \to & X \newline \downarrow & & \downarrow \newline Y & \to & Z. \end{array} $$

Note that $E$ need not be path-connected.

Is it possible to express $\pi_1(E,e)$ for a given choice of base point $e\in E$, in terms of $f_\sharp: \pi_1(X,x)\to \pi_1(Z,z)$ and $g_\sharp: \pi_1(Y,y)\to \pi_1(Z,z)$?

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  • $\begingroup$ I retract my previous thought... It seems like in general you'd need to know about $\pi_1$ of the fiber as well. Indeed, $S^\infty \times_{\mathbb{C}P^\infty} S^\infty$ is like $S^1$, so that seems to reduce hope. $\endgroup$
    – Dylan Wilson
    Commented May 30, 2013 at 14:24

3 Answers 3

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There is a ``Mayer--Vietoris" sequence $$\cdots \to \pi_2(Z, z) \to \pi_1(E, e) \to \pi_1(X, x) \times \pi_1(Y, y) \to \pi_1(Z,z) \to \pi_0(E) \to \cdots$$ that can be developed by fitting together the 4 long exact sequences of homotopy groups obtained from the 4 maps in your diagram.

With $X \simeq Y \simeq *$ and $Z=S^2$, $E \simeq \Omega S^2$ has fundamental group $\mathbb{Z}$, whereas $X$, $Y$ and $Z$ are simply-connected, so you can't express $\pi_1(E)$ solely in terms of the fundamental groups of the other spaces.

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    $\begingroup$ That sequence is quite useful, for some reason it usually doesn't appear in algebraic topology books. $\endgroup$
    – K.J. Moi
    Commented May 30, 2013 at 14:11
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    $\begingroup$ This is great, thanks. So are the maps in this sequence what you'd expect? In fact, I can see an obvious candidate for the map $\pi_1(E)\to \pi_1(X)\times \pi_1(Y)$, but not so much for the map $\pi_1(X)\times\pi_1(Y)\to \pi_1(Z)$. I want to send $(a,b)$ to $f_\sharp(a)\cdot g_\sharp(b)^{-1}$, but this won't be a homomorphism if $\pi_1(Z)$ isn't commutative. $\endgroup$
    – Mark Grant
    Commented May 30, 2013 at 14:59
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    $\begingroup$ @Mark: I don't know a reference. I hadn't noticed, but the map $\pi_1(X,x) \times \pi_1(Y, y) \to \pi_1(Z, z)$ is not a homomorphism (it is given by the formula you wrote), and exactness at this point is in the sense of pointed sets. Note that the preimage of the identity under that map is the same as the pullback of $\pi_1(X,x) \to \pi_1(Z,z) \rightarrow \pi_1(Y,y)$, so is a group. $\endgroup$
    – Oscar Randal-Williams
    Commented May 30, 2013 at 15:20
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    $\begingroup$ I agree that this should be in algebraic topology textbooks. The underlying algebra at least can be found in my book as Exercise 38 for section 2.2, although there is an implicit assumption there that all the groups are abelian. With just a little care the proof (diagram chasing) also works in the nonabelian case, with the caveat mentioned in earlier comments that the map from the product may not be a homomorphism. The exercise says in essence that the two long exact sequences of homotopy groups for the given fibration and its pullback combine to give Oscar's exact sequence. $\endgroup$
    – Allen Hatcher
    Commented May 30, 2013 at 18:14
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    $\begingroup$ This sequence is best (from my point of view) derived from the "Mayer-Vietoris fiber sequence" $\Omega Z \to E\to X\cross Y$ associated to the (homotopy) pullback square. This is in my book. $\endgroup$
    – Jeff Strom
    Commented May 31, 2013 at 1:17
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Expressed in terms of the homotopy pullback $N(f,g)$ of a pair of based maps $f\colon X\longrightarrow A$ and $g\colon Y\longrightarrow A$, the long exact sequence is Corollary 2.2.3 of May and Ponto ``More concise algebraic topology''. The result of which it is a corollary, Proposition 2.2.2, describes the pointed set of based maps $[Z,N(f,g)]$ for any based space $Z$. The dual result for homotopy pushouts is Proposition 2.1.2. The corollary is used heavily in the study of fracture theorems for localization and completion later on in the book.

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    $\begingroup$ I've accepted Oscar's answer as it came first, but this is the reference I was looking for, thank you. In particular I now see that Oscar's exact sequence is nothing other than the homotopy exact sequence of the fibration $E\to X\times Y$ which is the homotopy pullback of the diagonal map of $Z$ under $f\times g:X\times Y\to Z\times Z$. $\endgroup$
    – Mark Grant
    Commented May 31, 2013 at 8:35
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    $\begingroup$ And I now see that this is what @Jeff Strom was getting at in his comment above. $\endgroup$
    – Mark Grant
    Commented May 31, 2013 at 8:40
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Have a look at

(R. Brown, P.R. Heath and H. Kamps), "Groupoids and the Mayer-Vietoris sequence", J. Pure Appl. Alg. 30 (1983) 109-129.

A Mayer-Vietoris sequence for a pullback of a covering mao also appears in Section 10.7 of Topology and Groupoids, and was in the 1988 (differently named) edition.

Edit: Here is an extract from the above paper

(source)

which shows that there is some more information from the sequence than just the usual exact sequence. This sequence applies to spaces as is shown in Section 4 of the above paper. The point is that this detailed exactness is easier to extract in the groupoid model than directly in the topology.

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