I have a problem with writing rigorous solutions for the following exercises:
Let $S_g$ be an orientable surface of genus $g$. Calculate $H_*(S_g)$ using retractions. Hint: for example, $S_1 = T^2$ retracts onto a circle and if $g>1$, then $S_g$ retracts onto $S_1$ minus a small disk.
Let $N_g$ be a non-orientable surface of genus $g$, i.e., the connected sum of $g$ projective planes. Show that if $g>1$, then $N_g$ retracts onto the Mobius strip. Deduce a formula for the homology of $N_g$ in terms of the homology of $N_{g-1}$ (for $g>1$).
First of all, I can't clearly understand why torus retracts onto only one circle and not onto the wedge of two? Since torus is the surface of revolution generated by rotation one circle about another it seems to me that it may retracts onto the wedge of two circles. Similarly, I can't see why $S_g$ retracts onto $S_1\setminus D^2$.
Assuming that, we have $H_*(T^2) = H_*(S^1) \oplus H_*(T^2,S^1)$, but how to find homology of pair $(T^2,S^1)$? I know the final answer for torus but want to understand how it appears in the mentioned way.
As for the second problem, I know that $\mathbb{R}P^2$ is a square (which is homotopy equivalent to a disk) with glued Mobius strip. I think that even for $g=1$ we can retract $N_1$ onto the Mobius strip since the disk is contractible. If so, then $H_*(N_1) = H_*(S^1) \oplus H_*(\{pt\})$, where I wrote $S^1$ because Mobius strip is homotopy equivalent to an ordinary circle. Similarly, it seems to me that $N_g$ for arbitrarily $g$ retracts onto a wedge of $g$ Mobius strips (circles).
Can anyone explain me how should it be done correctly? Any help and hints will be very appreciative.
Best Answer
The retraction (not deformation retraction) of a torus onto only one circle can be seen geometrically by considering a torus which has the two circles intersecting $\mathbb{R}^2$ being a circle with radius $1$ and one with radius $2$. Take the projection $\pi: \mathbb{R}^3 \to \mathbb{R}^2$ on the first two coordinates, and the map $p:\mathbb{R}^2 -\{0\} \to S^1$ given by $p(x)=x/\Vert x\Vert$. Then $p \circ \pi $ is a retraction onto the circle. One more elementary way is just to take the projection $\pi: S^1 \times S^1 \to S^1$ on the first coordinate.
The wedge of two circles can't be a retract of the torus, since its fundamental group is nonabelian, and hence can't be a subgroup of $\mathbb{Z} \oplus \mathbb{Z}$.
Furthermore, we have that $(T^2,S^1)$ forms a "good pair". Note that $T^2/S^1 \sim S^2 \vee S^1$ (homotopy equivalent), which gives everything you want to calculate the homology of $T^2$.
For the $S_g$ part, consider the following image. I'll leave to you to understand what is the retraction.
(the case of $g>1$ is just collapsing everything beyond a certain height to a point).
For the second question, a similar argument used to prove that the wedge of two circles isn't a retract of the torus can be used to show that the mobius strip can't be a retract of $\mathbb{R}P^2$ as you suggest, since $\mathbb{Z}$ is not a subgroup of $\mathbb{Z}/2\mathbb{Z}$. Now, since $\mathbb{R}P^2$ is a mobius strip attached with a disk, if you remove that disk to make the connected sum, you get two mobius strips which you can simply project one onto the other (similarly to what happened above) to yield a retraction from $N_1$ to a mobius strip. The case of $g>2$ is similar.