Does homotopy type of a map $S^1 \to RP^1$ defines a homotopy type of $(D^2, S^1)\to (RP^2, RP^1)$?

Question

Consider a CW structure on $\mathbb{R}P^2$ with one 2-cell, one 1-cell $\mathbb{R}P^1$, and one 0-cell. Let $f \colon (D^2, \partial D^2) \to (\mathbb{R}P^2, \mathbb{R}P^1)$ be a map such that $f \big|_{\partial D^2}$ is homotopic to the attaching map $\phi \colon \partial D^2 \to \mathbb{R}P^1$ of the 2-cell. Does this imply that $f$ is homotopic to the characteristic map $i \colon (D^2, \partial D^2) \to \mathbb{R}P^2$ of the 2-cell, as maps of pairs?

Answer 1

$\newcommand{\RP}{\mathbb{R}P}$ $\newcommand{\Z}{\mathbb{Z}}$

I believe the answer is no.

Let's focus on the map $\pi_2(\RP^2, \RP^1) \to \pi_1(\RP^1)$ in the long exact sequence of homotopy groups for the pair $(\RP^2, \RP^1)$. If this map is injective, it implies that the homotopy class of a map $(D^2, S^1, x_0) \to (\RP^2, \RP^1, y_0)$ is uniquely determined by the homotopy class of its restriction $(S^1, x_0) \to (\RP^1, y_0)$.

Now, consider the long exact sequence of homotopy groups for $(\RP^2, \RP^1)$: $$ \dots \to \pi_2(\RP^2) \cong \Z \to \pi_2(\RP^2, \RP^1) \xrightarrow{j} \pi_1(\RP^1) \cong \Z \xrightarrow{t} \pi_1(\RP^2) \cong \Z_2 \to \dots $$

Note that the kernel of $t$ is $2\Z$. Consider the short exact sequence (SES): $$ 0 \to \Z \xrightarrow{l} \pi_2(\RP^2, \RP^1) \xrightarrow{j} 2\Z \cong \Z \to 0 $$

Since we can define $u \colon \Z \to \pi_2(\RP^2, \RP^1)$ such that $ju = \mathrm{id}_\Z$ by taking any preimage of a generator of the last $\Z$, the SES splits. Therefore, $\pi_2(\RP^2, \RP^1) \cong \Z \oplus \Z$, and $j$ is not injective.

It's important to note that this reasoning provides an answer for a pointed version of the question, considering maps between pointed spaces and homotopies preserving basepoints. Nevertheless, it is plausible that the unpointed version can be reduced to the pointed one.

(I would like to thank @Thorgott for pointing out mistakes in the first version of this answer and for several helpful comments)