Note: this is based on a problem in Hatcher (2.2.30) but is not the same problem. I'm just trying to understand how to use what I did there in different cases.
For a map $f:X \rightarrow X$, define the mapping torus of $X$ to be $X \times [0,1]/\sim$, where the equivalence $\sim$ identifies $(x,0) \sim (f(x),1)$. For $X = \mathbb{R} P^2 \vee \mathbb{R} P^2$, let $f:X \rightarrow X$ be the map that flips these two copies of $\mathbb{R}P^2$ i.e. sends a point in one copy of $\mathbb{R}P^2$ to its corresponding point in the other copy and vice versa. I'm trying to compute the fundamental group of this mapping torus. (And I'm just assuming the map preserves basepoints).
In the Hatcher problem (1.2.11 page 53), $X = S^1 \vee S^1$ and $f:X \rightarrow X$ is arbitrary. My argument there is that the mapping torus $T_f$ takes $S^1 \vee S^1 \times I$ and identifies $S^1 \vee S^1 \times \{0\}$ with $S^1 \vee S^1 \times \{1\}$ by the map $f$, and (since this map is assumed to be basepoint preserving) this means $I$ becomes a loop homeomorphic to $S^1$. So to get $T_f$, we can start with $S^1 \vee S^1 \vee S^1$ (where the last $S^1$ is from the loop of $I$) and attach the remaining copy of $S^1 \vee S^1$. Since $S^1 \vee S^1$ is one 0-cell and two 2-cells, this means we just need to attach two 2-cells to $S^1 \vee S^1 \vee S^1$ to build $T_f$.
So let $a,b$ be the generators of each $S^1$ in $S^1 \vee S^1$, and let $c$ be a generator for the loop from $I$. Then we can attach the first 2-cell by wrapping its edges first around $a$, then going along $c$ back to the basepoint on $X × \{1\}$, and gluing itself back to $X × \{0\}$ via the map $f_∗(a)^{-1}$, finally going back along $c^{-1}$ to where we started. We wrap the second 2-cell in basically the same way, now going along $b$ and $c$ instead of $a$ and $c$. By Van Kampen, $$\pi_1(T_f) = \langle a,b, c \mid ac f_*(a)^{-1} c^{-1}, a b f_*(b)^{-1}c^{-1}\rangle.$$ The problem is that I don't fully understand how to do this when we don't just have circles i.e. for $X = \mathbb{R} P^2 \vee \mathbb{R} P^2$. I know the idea of building $T_f$ from $X \vee S^1$ by attaching cells still makes sense, but the cell structure of $\mathbb{R} P^2 \vee \mathbb{R}P^2$ isn't as simple -- $\pi_1(\mathbb{R} P^2)$ has presentation $\pi_1(\mathbb{R} P^2) = \langle a,b \mid a^2 b^2 \rangle \simeq \mathbb{Z}_2 * \mathbb{Z}_2$ and each $\mathbb{R}P^2$ has the cell structure of one 0-cell, one 1-cell, and one 2-cell. What should I do here?
Update: My thinking is that we again build up $T_f$ from $\mathbb{R}P^2 \vee \mathbb{R}P^2 \vee S^1$ by attaching the cells of the other copy of $\mathbb{R}P^2 \vee \mathbb{R}P^2$. So we need to attach two 1-cells and two 2-cells. Let $a,b$ be the generators of $\mathbb{R}P^2 \vee \mathbb{R}P^2$ and $c$ be the generator of the loop formed from $I$. That is, we attach one 2-cell along $acf_*(a)^{−1}c^{−1}$ and the other along $bcf_∗(b)^{−1}c^{-1}$. But then my confusion is how should the two 1-cells be attached? Should I just glue them in the exact same way as gluing the edges of the 2-cells (i.e. one 1-cell along $acf_*(a)^{−1}c^{−1}$ and the other along $bcf_∗(b)^{−1}c^{-1}$)? Does this mean the 1-cells don't influence the computation under Van Kampen? Should the 1-cells be glued in some less complicated way? Thank you.