I am working on problems in the past Qual exams.
"Let $X$ be the space obtained from the 3-sphere $S^3$ by
identifying antipodal points on the equator $S^2$. Compute
$\pi_1(X)$."
I think about it like this: Let $A,B$ be the upper and lower hemispheres. Then $A=B=\mathbb{R}P^3$ and $C=A\cup B= \mathbb{R}P^2$. Set $\pi_1(A)=\langle a|a^2 \rangle,\pi_1(B)=\langle b|b^2 \rangle,\pi_1(C)=\langle c|c^2 \rangle$.
By Seifert-van Kampen theorem, our wanted group $G$ is the pushout of the diagram $f:C\to A$ and $g:C\to B$. The problem is, what is the maps $f$ and $g$? They are induced by inclusions, but I just found out that the induced maps are not necessarily injective. I tried to look at it geometrically, i.e. what is a loop in $C$ when considered in $A$? I believe that a loop $c$ when considered in $A$ is still a loop. But this is just an intuition.\
I did a homework problem about this, and the first homology of this space is $\mathbb{Z}_2$. By Huerwitz theorem, $f,g$ cannot be trivial maps.So my observation above must be right. But how do I provide a rigorous proof?
Best Answer
Let's focus on $f$ (of course for $g$ it's perfectly symmetric) : it is the inclusion at the equator of $\mathbb RP^2$ in $\mathbb RP^3$, so in fact it is induced by the inclusion $S^2\to S^3$ which is compatible with the antipodal action.
Now the goal is to recall how you proved that the $\pi_1$ of those spaces was $\mathbb Z/2$ : you (probably) used covering theory.
Now we have a commutative diagram $$\require{AMScd}\begin{CD} S^2 @>>> S^3 \\ @VVV @VVV \\ \mathbb RP^2 @>>> \mathbb RP^3 \end{CD}$$
Where each of the vertical maps is a $2$-sheeted covering. Now note that the top horizontal map is bijective on the fibers of the vertical covering maps : this means the following : if you take a point $x\in\mathbb RP^2$, and look at $p^{-1}(x)\subset S^2$, and then look at $x$ as a point in $\mathbb RP^3$ and $q^{-1}(x)\subset S^3$ ($p$ and $q$ are the vertical maps); well the top horizontal map induces a map $p^{-1}(x)\to q^{-1}(x)$: this map is a bijection.
Finally if you unravel the correspondance between covering spaces amd fundamental groups, you'll see that this bijectivity on fibers actually implies that the map is a bijection on $\pi_1$, so this tells you what $\pi_1(f)$ is (and similarly for $g$)
(More precisely: let $\gamma$ be a loop in $\mathbb RP^2$, and lift it to a path in $S^2$. Then we may see that as a path in $S^3$. Then by bijectivity in fibers, this path in $S^3$ is a loop if and only if it was one in $S^2$, so $\gamma$ is nullhomotopic in $\mathbb RP^3$ if and only if it was in $\mathbb RP^2$
This covers injectivity, which is of course sufficient in our case, but just for the sake of completeness and for more general cases : let $\gamma$ be a loop in $\mathbb RP^3$, then we may lift it to a path in $S^3$, between two points of the fiber. Then those two points have antecedents in the fiber in $S^2$, and any path between those in $S^2$ gets pushed to a homotopic path in $S^3$ (because $S^3$ is simply-connected, so the homotopy class of a path is entirely determined by its endpoints). We may thus, up to homotopy, assume the path comes from $S^2$, and so from $\mathbb RP^2$, thus ensuring surjectivity.)