Show that every continuous map \(f: P^2 \to S^1\) is nulhomotopic.
Find a continuous map of the torus into \(S^1\) that is not nulhomotopic.
Solution
Goal. (a) Every map \(P^2 \to S^1\) is null-homotopic. (b) Find a non-null-homotopic map \(T^2 \to S^1\).
1. (a) Every map \(f: P^2 \to S^1\) is null-homotopic.
1.1. \(f\) induces \(f_*: \pi_1(P^2) = \ZZ/2 \to \pi_1(S^1) = \ZZ\).
\(\pi_1(P^2) = \ZZ/2\) and \(\pi_1(S^1) = \ZZ\).
1.2. \(f_* = 0\).
the only homomorphism \(\ZZ/2 \to \ZZ\) is the zero map (there is no element of order \(2\) in \(\ZZ\)).
1.3. Hence \(f\) lifts to the universal cover \(\RR \to S^1\).
the lifting criterion: \(f_*(\pi_1(P^2)) = 0 \subseteq \pi_1(\RR) = 0\), so \(f\) lifts to \(\tilde f: P^2 \to \RR\).
1.4. \(\tilde f\) is null-homotopic (since \(\RR\) is contractible).
\(\RR\) is contractible, so any map into it is null-homotopic.
1.5. Hence \(f = p \circ \tilde f\) is null-homotopic.
composing a null-homotopy of \(\tilde f\) with the covering map \(p\) gives a null-homotopy of \(f\).
2. (b) A non-null-homotopic map \(T^2 \to S^1\).
2.1. Take the projection \(f: T^2 = S^1 \times S^1 \to S^1\) onto the first factor.
\(f(x, y) = x\).
2.2. \(f_*: \pi_1(T^2) = \ZZ \oplus \ZZ \to \pi_1(S^1) = \ZZ\) is the projection onto the first factor.
the induced map on \(\pi_1\) of a product projection is the projection.
2.3. \(f_*\) is nonzero (it is surjective), so \(f\) is not null-homotopic.
a null-homotopic map induces the zero map on \(\pi_1\).
3. Q.E.D.
1 proves (a); 2 gives the example for (b).