Computing homology of square with all vertices identified.

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.

• 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. – Qiaochu Yuan Nov 28 '15 at 23:24
• 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. – Ivo Terek Nov 28 '15 at 23:27
• 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.. – Ivo Terek Nov 28 '15 at 23:32
• Contract one of the edges. – user98602 Nov 28 '15 at 23:55
• Ah, ok! Thanks. – Ivo Terek Nov 28 '15 at 23:59

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 $\pi_1(S^1)$.