I'm trying to compute the homology of $X = (I \times I)/\sim$, where $(0,0)\sim (0,1) \sim (1,0) \sim (1,1)$. I want to do this via cellular homology, using degrees, etc, but I don't got that very well.

It is clear to me that we start with one $0$-cell, then glue four $1$-cells, then one $2$-cell, so the chain complex is: $$0 \to \Bbb Z \stackrel{d_2}{\to}\Bbb Z \oplus \Bbb Z \oplus \Bbb Z \oplus \Bbb Z \stackrel{d_1}{\to} \Bbb Z \to 0$$

If we call the vertex $v$, the $1$-cells $a, b, c$ and $d$, and $A$ the $2$-cell, then we have: $$d_2(A) = \deg(f_a) a + \deg(f_b)b + \deg(f_c)d + \deg(f_d)d,$$where each $f_i$ is the following map: $$\Bbb S^1\underbrace{\longrightarrow}_{\text{attaches $A$ to $X_1$}} X_1 \underbrace{\longrightarrow}_{\text{collapses $X_1$ except $i$}}\Bbb S^1$$

I'm fairly sure that all the degrees will be the same, so it suffices to compute $\deg(f_a)$. I have no idea of how to do that. I suspect that they'll be $1$, but I'm "thinking simplicially" here.

I mean, I don't know how to translate that collapsing of $X_1$ in terms of the word $abcd$ that represents the attaching map (thinking two more minutes the attaching map seems to be the identity, but I'm not sure at all of what I'm doing). Can someone help me? Thanks.

Edit: This specific problem is solved in the comments thanks to Qiaochu's nice observation, but I still don't get the bigger picture of these computations with degree vs words, so any explanation is welcome. This space can be used as an example yet.

  • 4
    $\begingroup$ I think it's cleaner to perform a deformation retract instead. You can think of this space as a "parachute." It deformation retracts onto the space obtained by connecting two points with four edges, which is a wedge of $3$ circles. So $H_0 = \mathbb{Z}, H_1 = \mathbb{Z}^3$, and all other homology vanishes. $\endgroup$ Commented Nov 28, 2015 at 23:24
  • $\begingroup$ I can see the parachute, but where did the $2$-cell went when you deformed to the wedge of circles? I'm sorry if my geometric vision here is too bad. $\endgroup$
    – Ivo Terek
    Commented Nov 28, 2015 at 23:27
  • $\begingroup$ Nevermind, I saw it. I just didn't saw how the space connecting two points with four edges is a wedge of three circles yet.. $\endgroup$
    – Ivo Terek
    Commented Nov 28, 2015 at 23:32
  • 1
    $\begingroup$ Contract one of the edges. $\endgroup$
    – user98602
    Commented Nov 28, 2015 at 23:55
  • $\begingroup$ Ah, ok! Thanks. $\endgroup$
    – Ivo Terek
    Commented Nov 28, 2015 at 23:59

2 Answers 2


Allow me to propose a different solution, using the Mayer-Vietoris sequence.

Take as open $U$ the space $X$ with the identified vertex deleted. This can be embedded as a convex subspace of $\mathbb{R}^2$ (a square) and so $H_n(U) = \delta_{n0} \mathbb{Z}.$ Next, as $V$, take a (small) open neighbourhood of the vertex. This space looks like a nuclear hazard symbol but with four triangles. A bouquet of four triangles with one vertex in common, say. This space can be embedded in $\mathbb{R}^2$ as a star-convex shape and so again $H_n(V) = \delta_{n0} \mathbb{Z}.$ The intersection $U \cap V$ is simply $V$ without the vertex, so a space consisting of four path connected triangles: $H_n(U \cap V) = \delta_{n0} \mathbb{Z}^4.$

Hence, for $n \geq 2$ we easily get from the Mayer-Vietoris sequence that $$0 \to H_n(X) \to 0$$ is exact, so $H_n(X) = 0$ when $n \geq 2$. The space is path connected so also $H_0(X) = \mathbb{Z}$.

Now let's compute $H_1(X)$. The relevant part of the Mayer-Vietoris sequence is $$H_1(U) \oplus H_1(V) \to H_1(X) \stackrel{\varphi}{\to} H_{0} (U \cap V) \stackrel{\psi}{\to} H_0(U) \oplus H_0(V),$$ which simplifies to $$0 \to H_1(X) \to \mathbb{Z}^4 \to \mathbb{Z} \oplus \mathbb{Z}.$$

Hence $\varphi$ is injective, so $H_1(X) \cong \text{im } \varphi = \ker \psi.$ But $H_0(U \cap V) \cong \mathbb{Z}^4$ is generated by a set of four points $x_1, x_2, x_3, x_4$, one chosen in each path component. So $$\psi(n_1 x_1 + n_2 x_2 + n_3 x_3 + n_4 x_4) = ((n_1 + n_2 + n_3 + n_4) x , -(n_1 + n_2 + n_3+n_4) x)$$ as these points all correspond to the same point $x$ in $H_0(U)$ and $H_0(V)$ (both infinite cyclic). Hence $$\ker \psi \cong \{ (n_1, n_2, n_3, n_4) \in \mathbb{Z}^4 \mid n_1 + n_2 + n_3 + n_4 = 0 \} \cong \mathbb{Z}^3.$$

It follows that $H_1(X) \cong \mathbb{Z}^3$. Remark that this is easy to generalize to a regular $n$-gon where we identify the $n$ vertices!


You can write the map $f_a$ down very explicitly. Namely, thinking of $S^1$ as $[0,1]/\{0,1\}$, $f_a$ is given by $f_a(t)=4t$ if $0\leq t\leq 1/4$ and $f_a(t)=0$ otherwise. Indeed, writing $f$ as the composition of two maps as you have done, the first map is the map that traverses loop $a$ on $[0,1/4]$, traverses loop $b$ on $[1/4,1/2]$, traverses loop $c$ on $[1/2,3/4]$, and traverses loop $d$ on $[3/4,1]$. The second map then just maps all the points in loops $b$, $c$, and $d$ to the basepoint, giving the map $f_a$ described above.

We can now just explicitly write a homotopy from $f_a$ to the identity to conclude that $f_a$ has degree $1$. For instance, we could take $H(t,s)=(4-3s)t$ for $t\in[0,1/(4-3s)]$ and $H(t,s)=0$ otherwise. When $s=0$, this gives $f_a$, and when $s=1$, it gives the identity.

In the same way, $f_b$, $f_c$, and $f_d$ are also homotopic to the identity. A bit more systematically, you can say that the attaching map $f:S^1\to X_1$ corresponds to the word $abcd$ in the free group $\pi_1(X_1)$ on four generators. The quotient maps $X_1\to S^1$ for each circle then just kill all but one of the generators, since all the circles but one are collapsed to the basepoint. So for instance, the quotient map for the second circle maps the word $abcd$ to $1b11=b$, which is just the homotopy class of the identity map in $\pi_1(S^1)$.


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