Here is the problem :
show that the map $f: \mathbb{R}P^2 \vee\mathbb{R}P^2 \rightarrow S^1$ is contractible.
I know the definition of contractible is that it is a space having homotopy type of a point. but I am still stuck about the proof of this, could anyone help me in this please?
EDIT:
Is the answer of my question similar to this one? Map to $RP^2 \vee S^1$ nullhomotopic
First, note that maps are not "contractible" as we reserve that word for spaces; the idea of a map being "zero" homotopically is called nullhomotopic. So we want to show that any map $f: \mathbb{R}P^2 \vee \mathbb{R}P^2 \to S^1$ is nullhomotopic.
The strategy for these kinds of arguments (if you can't just produce a nullhomotopy) is usually to show that $f$ factors through a contractible space. The first thing we could try is to use the lifting property for the universal covering map $p: \mathbb{R} \to S^1$, since $\mathbb{R}$ is contractible. So we want to know if there exists a map $\tilde{f}: \mathbb{R}P^2 \vee \mathbb{R}P^2 \to \mathbb{R}$ such that $p \circ \tilde{f} = f$.
To do this, we can use the lifting criterion, which says that this map $\tilde{f}$ exists iff $f_*(\pi_1(\mathbb{R}P^2 \vee \mathbb{R}P^2)) \subseteq p_*(\pi_1(\mathbb{R}))$ (I've suppressed basepoints), where $f_*$ and $p_*$ are the induced maps on fundamental groups. By van Kampen we know that $\pi_1(\mathbb{R}P^2 \vee \mathbb{R}P^2) \cong \mathbb{Z}_2 *\mathbb{Z}_2$, and since $\mathbb{R}$ is contractible, $\pi_1(\mathbb{R}) \cong 0$.
Now this means that $p_*(\pi_1(\mathbb{R})) \cong 0$, so by above, $\tilde{f}$ exists iff $f_*: \mathbb{Z}_2 *\mathbb{Z}_2 \to \pi_1(S^1) \cong \mathbb{Z}$ is zero (otherwise it can't have image contained in $p_*(\pi_1(\mathbb{R})) \cong 0$).
$f_*$ is a group homomorphism. But any homomorphism $\varphi: \mathbb{Z}_2 *\mathbb{Z}_2 \to \mathbb{Z}$ is necessarily zero: If we call the generators of $\mathbb{Z}_2 *\mathbb{Z}_2$ $a$ and $b$, then since $a^2 = 0$ in $\mathbb{Z}_2$ we have $0 = \varphi(a^2) = \varphi(a)^2 \implies \varphi(a) = 0$, and similarly for $b$. So $f_*$ is zero, and a lift $\tilde{f}$ of $f$ exists by the lifting criterion.
Thus $f: \mathbb{R}P^2 \vee \mathbb{R}P^2 \to S^1$ factors as the composition $\mathbb{R}P^2 \vee \mathbb{R}P^2 \xrightarrow{\tilde{f}} \mathbb{R} \xrightarrow{p} S^1$. But $\mathbb{R}$ is contractible, so $f$ is nullhomotopic. More explicitly, since $\mathbb{R}$ is contractible there is a homotopy $h: \mathbb{R}P^2 \vee \mathbb{R}P^2 \times I \to \mathbb{R}$ from $\tilde{f}$ to $c$ where $c$ is a constant map. But then postcomposing with $p$, we get $p \circ h: \mathbb{R}P^2 \vee \mathbb{R}P^2 \times I \to S^1$ is a homotopy from $p \circ \tilde{f}$ to $p \circ c$. The former is $f$, and the latter is constant. Therefore $f$ is nullhomotopic.