How do I compute the fundamental group of the connected sum $X \mathop{\#} X$, where $X$ denotes the real projective plane?
I'd like to use Van Kampen's theorem, but I have trouble visualizing what this space looks like. Could anyone provide hints?
[Math] The fundamental group of the connected sum of two copies of the real projective plane
algebraic-topologyfundamental-groupsgeneral-topology
Related Solutions
The background to this question is discussed in this mathoverflow discussion. Since two points, say $a,b$, are to be identified, it is as well to start with a model of the whole situation, namely the fundamental groupoid $G=\pi_1(X,C)$ on the set $C$ consisting of the two points $a,b$. Since $X$, presumably the $2$-sphere, is simply connected, $G$ is isomorphic to the groupoid $\mathcal I$ which has two objects $0,1$ and exactly one arrow $\iota:0 \to 1$, as well as $\iota^{-1}: 1 \to 0$, and identities at $0,1$. When you identify $0,1$ of $\mathcal I$ to a single point in the category of groupoids you get the group $\mathbb Z$ of integers.
All this is explained in the book Topology and Groupoids, the third edition of a book published in 1968, i.e. nearly 50 years ago. A translation of some 1984 comments of Grothendieck on the neglect of this use of many base points is given here. See also the account by Philip Higgins in this now downloadable Categories and Groupoids (originallly 1971).
My understanding is that algebraic topology deals with algebraic modelling of geometric situations, and groupoids, which may have many objects, allow the modelling of the identification of several base points. If you are tethered to only one base point, then things are not so clear, and students may get confused.
May 27: Just to add some information. The key point is that is that the category $Gpd$ has a forgetful functor $Ob: Gpd \to Sets$ which is a "cofibration" of categories, meaning that if $G$ is a groupoid and $f: Ob(G) \to Y$ is a function then there is a groupoid $f_*(G)$ and morphism $U_f: G \to f_*(G)$ with a universal property that can be expressed in a pushout diagram of groupoids $$\begin{matrix} Ob(G) & \to & Y\\ \downarrow & & \downarrow\\ G & \to & f_*(G) \end{matrix} $$ The construction of $f_*(G)$ was introduced by Philip Higgins and includes that of free groups and free products of groups.
The relevance of this to the generalised van Kampen Theorem is given in Section 9.1 of Topology and Groupoids.
As a matter of clarity, I denote $(a,i)$ and $(b,i)$ by $a_i$ and $b_i$ for $i=1,2,3$, respectively.
Lets call $Z$ the quotient space obtained from $Y$ by identifying $b_1$ with $a_2$ and $b_2$ with $a_3$. First we can compute the fundamental group of $Z$. By applying Van Kampen twice, we get $$\pi_1(Z)\simeq\langle c_1,c_2,c_3\vert c_1^2=c_2^2=c_3^2=1\rangle,$$ where $c_i$ corresponds to the generator of the fundamental group of $RP^2\times \{i\}$.
Now $X$ is obtained from $Z$ by identifying $a_1$ and $b_3$. Let $$p:Z\to X$$ be the corresponding quotient map and let $\overline{a_1}=p(a_1)=p(b_3)$. The map $p$ induces a morphism $$p_*:\pi_1(Z,a_1)\longrightarrow \pi_1(X,\overline{a_1}).$$ We would be happy to say that $p_*$ is onto, but it's not. This is because if $\tilde{\gamma}:[0,1]\to Z$ is a path from $a_1$ to $b_3$, its image $\gamma=p(\tilde{\gamma})$ is a loop in $X$, but it won't be in the image of $p_*$. Let $t$ be some object (I just need $t$ to be some object generating a copy of $\Bbb Z$), there is a natural homomorphism $$f:\langle t\rangle\longrightarrow \pi_1(X,\overline{a_1})$$ sending $t$ to $[\gamma]$. Then $p_*$ and $f$ induce a morphism $$\psi:\pi_1(Z,a_1) * \langle t\rangle\longrightarrow \pi_1(X,\overline{a_1}).$$ I claim that $\psi$ is an isomorphism. If you admit this (which is far from trivial, but reasonable if you picture what $X$ looks like) you get $$\pi_1(X)\simeq\langle c_1,c_2,c_3,t~\vert~ c_1^2=c_2^2=c_3^2=1\rangle.$$
I know that $\psi$ is an isomorphism from a quotient version of Van Kampen that I learned here (unfortunately it is in French). In this reference they also call it the HNN-version of Van Kampen but I haven't found any reference to this specific theorem by searching with these words. If you want to prove that $\psi$ is an isomorphism, you can do something like this:
$\psi$ is onto: Take $U_1=X-\{\overline{a_1}\}$ and $U_2=p(V_1\cap V_2)$ where $V_1$ and $V_2$ are two open sets of $Z$ that deform retract onto $a_1$ and $b_3$, respectively. $U_1$ and $U_2$ are two open sets that cover $X$. Take a path $\alpha:[0,1]\to X$. Then you can decompose $\alpha$ as the concatenation $$\alpha=\alpha_1\cdot\alpha_2\cdots\alpha_{n-1}\cdot\alpha_n$$ such that $\alpha_i$ is a path in $U_1$ or in $U_2$. Then use the fact that $p:Z-\{a_1,b_3\}\to U_1$ is an homeomorphism to express each path $\alpha_i$ with image in $U_1$ as a combination of $\gamma$ and of elements of $\pi_1(Z)$. This kind of technique is very similar to the proof of the classical theorem of Van Kampen.
$\psi$ is one-to-one: You can use the covering of $X$ by $$\tilde{X}=RP^2\times \Bbb Z/_{b_i\sim a_{i+1}}.$$ Take some combination of $\gamma, c_1, c_2$ and $c_3$ which is trivial in $X$. You can picture what the lift will look like in $\tilde{X}$ in terms of the coefficient of the combination, and you should understand that the combination must be trivial because the lift is trivial in $\tilde{X}$.
I hope this helps! I didn't add much detail at the end but I can add some later if you want. I can add drawings to clarify the situation too if you'd like.
Best Answer
The real projective plane $\mathbb{RP}^2$ can be represented as the following identification bi-gon:
By taking two of these bigons, cutting each at a vertex, and gluing them together again, we obtain the identification polygon for $\mathbb{RP}^2 \# \mathbb{RP}^2$:*
Now, let $P$ be an interior point of the identification polygon. Then, we can apply the Seifert-van Kampen Theorem to $\mathbb{RP}^2 \# \mathbb{RP}^2 = \mathbb{RP}^2 \# \mathbb{RP}^2 \setminus \{P\} \cup D$, where $D$ is a small disk containing $P$. **
Can you take it from here?
* : Note that $\mathbb{RP}^2 \# \mathbb{RP}^2$ is in fact homeomorphic to the Klein bottle. This can be seen by cutting and regluing the identification polygon.
** : This is a standard trick for finding fundamental groups of compact surfaces. We first realize the surface as an identification polygon, and then apply the Seifert-van Kampen Theorem to the polygon with a point $P$.removed and a small disk around $P$.