I am having trouble following the argument in Massey for the proof of Theorem 8.3, which states that if $f: X \rightarrow Y$ is a homotopy equivalence, then $f_*: \pi(X, x) \rightarrow \pi(Y, f(x))$ is an isomorphism for any $x \in X$.

The proof in the book assumes Theorem 8.2., which states that if $\phi_0, \phi_1: X \rightarrow Y$ are continuous maps which are homotopic, then the image of $\pi(X, x_0)$ under $\phi_{0*}$ and $\phi_{1*}$ are isomorphic.

They use this to claim that $\pi(X, x) \rightarrow \pi(X, gf(x))$ factors through $\pi(Y, f(x))$ via $f_*$ and $g_*$. I do not see why this is true. It seems that theorem 8.2 guarantees that a different diagram commutes.

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    $\begingroup$ I think that's suppose to be $f_*\colon\pi(X,x)\to\pi(Y,f(x))$, no? Also: what "Theorem 8.3"? Massey wrote more than one book, and at least the one I have has several "Theorem 8.3". But it seems to be Theorem II.8.3, in page 57, of A Basic Course in Algebraic Topology. $\endgroup$ – Arturo Magidin Jan 21 '11 at 17:58

Apply Theorem II.8.2 with $Y=X$, $\varphi_0 = \mathrm{id}_X$, and $\varphi_1$; You get: $$\begin{array}{ccc} \pi(X,x) & \stackrel{\mathrm{\varphi}_{0*}}{\longrightarrow} & \pi(X,x)\\ \varphi_{1*}&\searrow & \downarrow u\\ &&\pi(X,gf(x)) \end{array}$$ Now, $\varphi_{0*}=\mathrm{id}_*$ is just the identity, and $\varphi_{1*} = (gf)_* = g_*f_*$, so you can rewrite the above as: $$\begin{array}{ccc} \pi(X,x) & \stackrel{\mathrm{id}}{\longrightarrow} & \pi(X,x)\\ g_*f_*&\searrow & \downarrow u\\ &&\pi(X,gf(x)) \end{array}$$ This means that $u = (gf)_* = g_*f_*$, which is exactly what the diagram at the bottom of page 57 says.


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