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Let $R$ be $S^{1}\vee S^{1}$. Call the first circle by $a$ and the second one by $b$. Let $X$ be space by attaching two $2$-cells to $R$ one via the boundary map $a^{3}$ and the other via the boundary map $ababababab$. Show that any map $f: X\to \mathbb{RP}^{2}\vee S^{1}$ is null-homotopic.

P.S: One can start by looking at the induced map between fundamental groups and then lift $f$ up to the universal cover, but then the problem is the universal cover of $\mathbb{RP}^{2}\vee S^{1}$ is not contractible and this does not work! Any other idea is welcome!

Peyman Morteza
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Let $\alpha$ be the generator of $\pi_1(\mathbb RP^2)$ and $\beta$ be the generator of $\pi_1(S^1)$. Note that $f_*(a)$ can be written only two ways: $\alpha\beta^{k_1}\alpha\dots\alpha\beta^{k_n}\alpha\,\,$ or $\,\,\beta^{k_1}\alpha\beta^{k_2}\dots\alpha\beta^{k_n}\,\,$ for some $k_i\in\mathbb Z$. Then, as it easy to see, if $k_1\ne -k_n$, we cannot reduce letters "β" so that $f_*(a)^3=1$; if $k_1=-k_n$, we can act on the $f_*$ by conjugation, and then the image of $a$ will became $\alpha\beta^{k_2}\dots\beta^{k_{n-1}}\alpha\,$ or $\,\beta^{k_2}\alpha\dots\alpha\beta^{k_{n-1}}$. Repeating this we can show that $f_*(a)=1$.

The same algorithm shows that $f_*(b)^5=1\,\Rightarrow\,f_*(b)=1$.

Andrey Ryabichev
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