Let \(i: \mathbb{RP}^2 \to \mathbb{RP}^3\), induced by \(S^2 \hookrightarrow S^3\) as the equator. Show that \(i \not\simeq \text{const}\).
Solution
1. The inclusion\[ i:\mathbb{RP}^2\hookrightarrow\mathbb{RP}^3 \]induces the identity isomorphism\[ i_*:\pi_1(\mathbb{RP}^2)\cong\mathbb Z/2\longrightarrow\pi_1(\mathbb{RP}^3)\cong\mathbb Z/2. \]
Use the standard CW structures \[ \mathbb{RP}^2=e^0\cup e^1\cup e^2, \qquad \mathbb{RP}^3=\mathbb{RP}^2\cup e^3. \] Attaching cells of dimension at least \(3\) does not change the fundamental group. Hence the inclusion of the \(2\)-skeleton induces an isomorphism on \(\pi_1\).
2. A null-homotopic map induces the trivial homomorphism on fundamental groups.
Homotopic maps induce the same map on \(\pi_1\) up to the usual basepoint conjugacy. A constant map induces the zero homomorphism, and conjugating the trivial homomorphism leaves it trivial.
3. Therefore \(i\) is not null-homotopic.
1. Hence\[ \boxed{i\not\simeq\mathrm{const}}. \]