I am just practicing how to use Seifert-van Kampen. The following excercise is from Hatcher, p.53.

Let $X$ be the quotient space of $S^2$ obtained by identifying the north and south poles to a single point. Put a cell complex structure on $X$ and use this to compute $\pi_1(X)$.

I found a cell complex structure for $S^2$ with two points $e^0_1$, $e^0_2$ and a line $e^1$, and a disc $e^2$. Then I identify both points, such that one remains. I choose open neighbourhoods around $e^1$ and $e^2$, let`s call them $U(e^1)$ and $U(e^2)$, then apply Seifert van Kampen with those as the subsets. Then just $\pi_1(U(e^1))$ is $\mathbb{Z}$, and the other group $\pi_1(U(e^2))$ is trivial.

So now the contributions from the intersection $U(e^1)\cap U(e^2)$ are trivial and we get $\mathbb{Z}$ for the fundamental group. Is this correct?


2 Answers 2


This question illustrates the compelling case to use groupoids and the fundamental groupoid on a set of base points. In particular we take the $2$-sphere $S^2$ and the North and South poles $n,s$, say. Let $A$ be the set consisting of $n,s$. Then the fundamental groupoid $G = \pi_1(S^2,A)$ has only one arrow $\iota: n \to s$.

Now we can talk about identifying $n,s$ in the groupoid $G$ to a point $p$, say. This gives a pushout of groupoids $$\begin{matrix} A & \to & \{p\}\\ \downarrow && \downarrow\\ G & \to & H \end{matrix}$$ (We regard $A$ and $\{p\}$ as the same as the sets of identities in the groupoids.) An easy universal argument gives that $H$ is isomorphic to $\mathbb Z$, the additive group of integers. That is, using groupoids means you immediately get a model of the topology.

Full details of this kind of argument are given in the book Topology and Groupoids, T&G, whose first edition with these ideas was in 1968.

Note that since groupoids can be seen as having structure in dimensions $0$ and $1$, we can consider identifications in dimension $0$, which gives additional flexibility to geometric and topological applications, such as that in the question. See for example this mathoverflow discussion.

Grothendieck wrote in his 1984 "Esquisse d'un Pogramme" (English translation):

" ..,people still obstinately persist, when calculating with fundamental groups, in fixing a single base point, instead of cleverly choosing a whole packet of points which is invariant under the symmetries of the situation, which thus get lost on the way. In certain situations (such as descent theorems for fundamental groups `a la van Kampen) it is much more elegant, even indispensable for understanding something, to work with fundamental groupoids with respect to a suitable packet of base points,.."

I conjecture that one reason for this neglect is that people are unaware of the techniques for computing with groupoids. One of the constructions developed by Philip Higgins in his book Categories and Groupoids is to consider for groupoids pushouts of the form $$\begin{matrix} X & \xrightarrow{f} & Y \\ \downarrow & & \downarrow \\ G & \to &H \end{matrix} $$ where $X=Ob (G), Y=Ob(H)$ and $f:X \to Y$ is a function, and the downarrows are essentially inclusions of identities. Higgins writes $H$ as $U_f(G)$ but a convenient notation is also $f_*(G)$. This is related to the groupoid van Kampen Theorem in T&G, 9.1.2 (Corollary 3), p.343. The explicit construction of $f_*(G)$ generalises the construction of free groups and of free products of groups.

Oct 12, 2015 Actually the question does ask about cell structures. The following Figure 7.9 from T&G should help:

modified sphere

The left figure shows the sphere with antipodal points identified. It is clear that this is a torus $T= S^1 \times S^1$ with an inner circle shrunk to a point. But $T$ has a cell decomposition as $e^0 \cup e^1 \cup e^1 \cup e^2$ and so the shrinking kills one of the $e^1$.

The right hand figure shows the "homotopy identification" $Y$ of $n,s$ by adding a line. That space $Y$ is clearly homotopy equivalent to $S^1 \vee S^2$.

For more background on groupoids in algebraic topology see also this presentation.


There is an answer to this problem on the wikipedia page for the theorem.

More generally, if you can recognize something as a wedge there is an easy way to apply the theorem -- the fundamental group is the free product of the wedge components:

"Van Kampen's theorem gives certain conditions (which are usually fulfilled for well-behaved spaces, such as CW complexes) under which the fundamental group of the wedge sum of two spaces X and Y is the free product of the fundamental groups of X and Y."

  • $\begingroup$ I just see the 2-sphere there with no points identified, which has trivial fundamental group $\endgroup$
    – w_w
    Oct 9, 2015 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.