Give an example of a homotopy class of maps of \(S^1 \lor S^1\) each member of which must have a fixed point, and also an example of a map of \(S^1 \lor S^1\) which doesn’t have a fixed point.
Solution
Let \[ X=S^1_a\vee S^1_b \] with wedge point \(v\).
1. Every self-map of \(X\) homotopic to the identity has a fixed point.
The graph \(X\) has one \(0\)-cell and two \(1\)-cells, so \[ H_0(X;\QQ)\cong\QQ, \qquad H_1(X;\QQ)\cong\QQ^2, \] and all higher homology groups vanish.
Let \[ g:X\longrightarrow X \] be homotopic to \(\operatorname{id}_X\). Homotopic maps induce the same maps on homology, hence \[ g_*=\operatorname{id} \] on both nonzero homology groups. Its Lefschetz number is therefore \[ L(g) =\operatorname{tr}(g_*|H_0)-\operatorname{tr}(g_*|H_1) =1-2 =-1. \] In particular, \[ L(g)\ne0. \] By the Lefschetz fixed-point theorem, \(g\) has a fixed point. Thus the homotopy class of the identity is an example in which every representative has a fixed point.
2. There is a fixed-point-free self-map of \(X\).
Identify the first circle \(S^1_a\) with the unit circle in \(\CC\) so that its wedge point is \[ v=1. \] Let \[ r:X\longrightarrow S^1_a \] be the retraction that is the identity on \(S^1_a\) and collapses the entire second circle \(S^1_b\) to \(v\). Let \[ R:S^1_a\longrightarrow S^1_a, \qquad R(z)=-z, \] be rotation by \(\pi\), and define \[ f=R\circ r:X\longrightarrow S^1_a\subseteq X. \]
If \(x\in S^1_a\), then \[ f(x)=-x\ne x. \] If \(x\in S^1_b\), then \[ f(x)=R(v)=-1. \] The point \(-1\) lies in \(S^1_a\setminus\{v\}\), while \[ S^1_a\cap S^1_b=\{v\}. \] Hence \(f(x)\ne x\) for every \(x\in S^1_b\), including \(x=v\).
Thus \(f\) has no fixed points.
Therefore the identity homotopy class supplies the first requested example, while the explicit map \[ \boxed{f=R\circ r} \] supplies the second.