I'm studying for an oral qualifying exam and going through various past exams I find on the interwebs, including this mock exam from the University of Bath. One of the questions seems like it should be rather straightforward, but it's giving me trouble. The question goes as follows:

Four small squares are cut out from a large square and discarded. The edges of what remains are then identified according to the labels in the figure. Find the fundamental group of the resulting space $X$. [You may wish to introduce more edges to help your calculations, but be careful not to make unauthorized identifications.]

The Space X

What would be a 'standard' approach to such a question? Here is what I have thought of so far and where I'm stuck:

  • Using Van Kampen's: Intuitively, I'd like to set $U$ to be the interior of the large square and $V$ to be the enlargement of the boundary of the large square. Then $U \cap V \simeq S^1$, but I get tangled up when it comes to making the edge identifications and determining what to do with $U$, plus, after edge identifications, $U$ and $V$ interact, and so $U \cap V$ might not actually be $S^1$ as I've thought it was. Is Van Kampen's the way to go here?

  • Since the hint mentions adding edges, is there some way to view this as a simplicial complex and compute the fundamental group from there? If so, how would I go about computing the fundamental group? (I am only familiar with computing homology for simplicial complexes, where I explicitly construct the boundary operators and compute the kernel/image of each to find the quotient groups.)

Thank you all for your help/advice!


New Approach #1: I'm going to try and approach this by using Proposition 1.26 in Hatcher. That is, I'll view $X$ as attaching a 2-cell to the boundary in the picture (as suggested in the comments). First, I've added a basepoint $x_0 \in X$ and edges from this basepoint to the points labeled $C$ in each of the squares, as shown in this picture:

Space with edge labels and added edges.

After doing this, the boundary of the space turns into the following picture, and, by contracting the edge $k$ to a point, this space deformation retracts to the wedge of five circles. I've also listed the attaching map.

Boundary of attached cell.

By the proposition, we should be able to compute $\pi_1(X) = \pi_1(\partial X)/N$ where $N = \langle e\gamma e^{-1}f \gamma f^{-1} g \gamma g^{-1} h \gamma h^{-1} k \gamma^{-1} k^{-1} \rangle$. Writing $\pi_1(\partial X) = \langle \gamma, e, f, g, h \rangle$ and noting that $k$ is now the identity element, we get $\pi_1(X) = \langle \gamma, e, f, g, h \rangle / \langle e\gamma e^{-1}f \gamma f^{-1} g \gamma g^{-1} h \gamma h^{-1} \gamma^{-1} \rangle$.

  • $\begingroup$ Try Van Kampen again, but with $U$ as the light gray region (ie all but the squares) and $V$ as a union of enlargements of each square. $\endgroup$ – aes Jan 1 '15 at 23:58
  • $\begingroup$ You can realize this space as a CW-complex by attaching one $2$-cell to a wedge sum of $5$ circles. The attaching map can be inferred from the diagram. This is one way to compute the fundamental group. $\endgroup$ – Ayman Hourieh Jan 2 '15 at 1:10
  • $\begingroup$ @aes - In order to use Van Kampen, we need $U \cap V$ to be path-connected, but the way I see it, with the $U$ and $V$ you've suggested, $U \cap V$ becomes five disjoint copies of $S^1$, and so it is not path-connected. Is there something I'm missing here with the identifications? $\endgroup$ – Rachel Jan 2 '15 at 19:00
  • $\begingroup$ @AymanHourieh - I've added details to trying to implement the approach you suggested. I'd be very grateful if you had a chance to look it over and offer any suggestions, as I wasn't able to complete the problem. Thank you! $\endgroup$ – Rachel Jan 2 '15 at 20:10
  • 3
    $\begingroup$ @Rachel: Instead of including your solution in the question, why not turn it into an answer? That way the question will no longer be on the unanswered list. $\endgroup$ – Michael Albanese Apr 17 '15 at 3:54

I'll give here another answer which i think is correct but I am not 100% sure. We will rely heavily on the first theorem at page 11 of Hatcher's Algebraic Topology, which basically allows us to do 2 things:

  1. We can kill any contractible line without changing homotopy type
  2. Instead of identifying two points we can just attach a $1$-cell at these two points.

Our first observation is that after identifying the edges, non of the vertices of the large square has been identifyied with eachother therefore we can collapse each edge of the edges $AB,BC,CD,DA$ to a point. The result space is Space $1$, where red points means identified (see the picture).

Now, we get the homotopy equivalent Space $2$ by replacing identifications by $1$-cells.

At last, we pull the outside point to the center, to get Space $3$, where the red lines are to be thought above the shape.

enter image description here

Now for the tricky part: The disc, without the red lines, deformation retracts to the wedge of $4$ circles. To picture this, imagine a strong gravity well at the center of the disk pulling everything radially at the center of the disc. Evertyhing either falls to the boundaries of the 4 existing holes or at the center of the disc, including the attaching points, therefore the red lines now became circles.

Therefore we end up wit the wedge of 8 circles and the fundamental group is $F_8$.


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