# The fundamental group of $B^2\times S^1$

In one exercise we are supposed to find the fundamental group of $B^2\times S^1$. It is given that the fundamental group is $\mathbb{Z}$, because we can show that $S^1$ is a deformation retract of this space.

The deformation retract they use is this: Look at the space a a doughnut, but it is a solid doughnut, not the standard-torus. Then if you cut the doughnut along it smaller circle,(so that you get a curved sausage), you can define a deformation retraction in this crosssection, by retracting everything to the center, and hence you get a deformation-retraction to $S^1$.

But what I am wondering is why we can not get that $B^2$ is a deformation-retraction? What I was thinking was that at each point inside the doughnut you rotate each point counterclockwise uintil you get to a the $B^2$ crosssection at your choosing. That is, each pont inside the doughnut travel counterclockwise uintil all the points get to a $B^2$-cross-section. Why is this not a deformation retraction? More formally if our point in $B^2\times S^1$ is $(x_1,x_2,y_1,y_2)$(this is not how it is visualized as a doughnut in $\mathbb{R}^3$), then we rotate it so that it becomes the point $(x_1,x_2,1,0)$

An equation for this roptation is: we have that $(y_1,y_2)=(\cos(2\pi s),\sin(2\pi s))$ for some s, our deformation retraction would be $(\cos(2\pi s(1-t) ),\sin(2\pi s(1-t))$.

Can you please exaplain why this does not work? It seems like we then get a homotopy between the identity map and $B^2$?

Maybe what goes wrong is that we have to work with s?, in the first case the homotpy is simply $(x_1,x_2,y_1,y_2)\rightarrow (tx_1,tx_2,y_1,y_2)$. In the second case it is $(x_1,x_2,y_1,y_2)\rightarrow(x_1,x_2,s)\rightarrow(x_1,x_2,\cos(2\pi s(1-t),\sin(2\pi s(1-t)))$.

Do yous see where the second approach is wrong?

• Sorry, but that the deformation retract to $S^1$ is not clear to me. You don't cut. To get it, you just compress the doughnut tightly. Commented Dec 29, 2015 at 18:09
• @AloizioMacedo Yes you are correct, the cutting part was just to visualize what happens at the cross-section, but your explanation is probably better. Commented Dec 29, 2015 at 18:13
• Oh, I see now what you meant. : ) Commented Dec 29, 2015 at 18:14

Mathematically, your homotopy is discontinuous at $t = 0$ in the disk corresponding to $s = 0$ (and $s = 2 \pi$).