Construct a connected graph \(X\) and maps \(f, g: X \to X\) such that \(fg = \mathbb{1}\) but \(f\) and \(g\) do not induce isomorphisms on \(\pi_1\). [Note that \(f_*g_* = \mathbb{1}\) implies that \(f_*\) is surjective and \(g_*\) is injective.]
Solution
Let \[ X=\bigvee_{n=0}^{\infty} C_n \] be the CW wedge of countably many circles at a common basepoint \(x_0\). Choose orientation-preserving homeomorphisms \[ h_n:C_n\to C_{n+1} \] fixing \(x_0\).
1. Define\[ g:X\to X \]by\[ g|_{C_n}=h_n \]for every \(n\ge0\).Define\[ f:X\to X \]by collapsing \(C_0\) to \(x_0\) and, for \(n\ge0\), setting\[ f|_{C_{n+1}}=h_n^{-1}. \]
Each map is continuous on every closed cell and fixes the common \(0\)-cell. By the weak topology of the CW wedge, the resulting maps are continuous on \(X\).
2. One has\[ f\circ g=\operatorname{id}_X. \]
For every \(n\ge0\), \[ (f\circ g)|_{C_n}=h_n^{-1}\circ h_n=\operatorname{id}_{C_n}. \] The wedgepoint is fixed as well.
3. The fundamental group is\[ \pi_1(X,x_0)\cong F(a_0,a_1,a_2,\dots), \]where \(a_n\) is represented by \(C_n\).The induced maps satisfy\[ g_*(a_n)=a_{n+1}, \]and\[ f_*(a_0)=1, \qquad f_*(a_{n+1})=a_n. \]
The fundamental group of a graph is free on the edges outside a maximal tree; here the \(0\)-skeleton itself is a maximal tree, so the circles give the stated free basis. The formulas follow directly from the definitions of \(f\) and \(g\) on each circle.
4. Neither \(f_*\) nor \(g_*\) is an isomorphism.
The map \(g_*\) is injective but not surjective, since no reduced word in its image contains the generator \(a_0\). The map \(f_*\) is surjective but not injective, since \[ a_0\ne1 \] while \[ f_*(a_0)=1. \] Nevertheless \[ f_*g_*=(fg)_*=\operatorname{id}. \] Thus \(f,g\) have the required properties.