Help with amalgated product (calculating $\pi_1(\mathbb{R}P^2)$)

Question

I’m trying to calculate $\pi_1(\mathbb{R}P^2)$ using Van Kampen’s theorem. After choosing the two open sets (which I call $U$ and $V$) according to my lecture notes, I get $$\pi_1(\mathbb{R}P^2)=\pi_1(S^1)*_{\pi_1(S^1)}\pi_1(D^2)=\mathbb{Z}*_{\mathbb{Z}}\{e\}$$

I know this has to give $\mathbb{Z}_2$ as a result, but I’m not sure how to apply the definition of the amalgamated product in order to find this (the definition is still a bit obscure for me, as I just learnt about it). I think I should take a generator of $\mathbb{Z}$ and look at its images by the induced homomorphisms $(i_1)_*$ and $(i_2)_*$, where $i_1:U \cap V \rightarrow U$ and $i_2:U \cap V \rightarrow V$ are the inclusions, and find something like the condition $\langle b \mid b^2=1 \rangle$, but I don’t see exactly how to do that.

Could someone please guide me through the steps?

Answer 1

You might find this formulation of the almagamated product (pushout) in terms of group presentations useful. The fundamental group $\pi_1(U \cup V)$ is constructed as the quotient of the free product of $\pi_1(U) * \pi_1(V)$ by the smallest normal subgroup that identifies the images of the generators of $\pi_1(U \cap V)$ in the groups $\pi_1(U)$ and $\pi_1(V)$.

In your example, the free product is $$ \pi_1(U) * \pi_1(V) = \langle\, b \mid - \,\rangle * \langle\, - \mid - \,\rangle = \langle\, b \mid - \,\rangle \cong \mathbb{Z}, $$ where we use the symbol $-$ to indicate an empty set of generators or relations. Now suppose that $a$ freely generates the fundamental group of the intersection, i.e. $\pi_1(U \cap V) = \langle\, a \mid - \,\rangle \cong \mathbb{Z}$. The induced maps are $(i_1)_*(a) = b^2 \in \pi_1(U)$ and $(i_2)_*(a) = 1 \in \pi_1(V)$. Hence, in the amalgamated product, we have the relation $b^2 = 1$, so $$ \pi_1(X) = \pi_1(U \cup V) = \langle\, b \mid b^2 \,\rangle \cong \mathbb{Z}_2. $$

Answer 2

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This is how you should think about it: $\mathbb{R}P^2$ is glued from $S^1$ and $D^2$ by identifying $S^1$ with $\partial D^2 = S^1$ via the map $S^1 \xrightarrow{\times 2} S^1.$ We can depict this as the following diagram:

\begin{CD} S^1 @>{incl}>> D^2 \\ @V{\times 2}VV @V{}VV \\ S^1 @>{}>> \mathbb{R}P^2. \end{CD}

This is what is called a pushout diagram - the point is precisely what is written above: the lower-right corner is glued from the upper-right and lower-left corners along the upper-left. [Strictly this means that you get $D^2 \coprod S^1$ and factor by the relation that for all $x$ from the upper-left $S^1$, their images in $D^2 \coprod S^1$ are identified.]

Now, the point of Van-Kampen is that the same will happen of the level of fundamental groups. That is, the diagram

\begin{CD} \pi_1(S^1) @>{\pi_1(incl)}>> \pi_1(D^2) \\ @V{\pi_1(\times 2)}VV @V{}VV \\ \pi_1(S^1) @>{}>> \pi_1(\mathbb{R}P^2). \end{CD}

is also a pushout, meaning that $\pi_1(\mathbb{R}P^2)$ is glued from $\pi_1(D^2) \simeq 1$ and $\pi_1(S^1) \simeq \mathbb{Z}$ and the relation is that images of elements of the upper-left $\pi_1(S^1)$ are identified. If the lower $\pi_1(S^1)$ is generated by some $a$ and the upper $\pi_1(S^1)$ is generated by $b$, the above means that we should take $\left<a \right> * 1$ and add the relation that $\pi_1(\times 2)(b) = \pi_1(incl)(b)$ inside this group. Explicitly, this is $"2a = e"$, because $\pi_1(\times 2)$ sends a generator to twice a generator, whereas $\pi_1(incl)$ of course just goes to the trivial group $1$. Hence the representation is $\left<a | 2a \right>. $