Computing homology of $S^1$ as in J.W. Vick's Homology Theory book and other questions

I am reading the book Homology Theory (GTM 145) by James W. Vick. On pages 25-26, the first homology of the circle is computed. However, in the computation, there is a sentence I do not understand.

On line 3 and 4 of page 26, it reads

Since $U$ and $V$ are pathwise connected, $i_{*}(ax+by)=0$ if and only if $a=-b$

I do not understand why this is so. It might be trivial, but please help. I have thought about it on and off for days.

Another questions is this

If we remove a point from the interior of a Möbius band, is the resulting space homotopy equivalent to a familiar space?

Thanks.

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It might be helpful to state what the symbols are. I'll try to pull up the relevant pages in Google Books, but that's always hit or miss. [Edit: And it appears that there are two editions. In the edition on Google Books, this is on page 23.] – Dylan Moreland Feb 26 '12 at 15:53
@Dylan, you are right. Thanks. Describing the symbols will make the question very long. That was why I only gave reference to the relevant page. – Herband Feb 26 '12 at 19:13

This answer is maybe more complete than you want, but hopefully it is beneficial to on-lookers:

Here $U$ is $S^1$ minus the south pole and $V$ is $S^1$ minus the north pole. He is using the Mayer-Vietoris sequence

$H_1(U)\oplus H_1(V)\rightarrow H_1(S^1)\rightarrow H_0(U\cap V)\stackrel{g_*}{\rightarrow} H_0(U)\oplus H_0(V)$

This is an exact sequence, meaning the image of one arrow is exactly the kernel of the next arrow.

The first term $H_1(U)\oplus H_1(V)$ is $0$ because $U$ and $V$ are contractible (so all but $H_0$ vanishes for each of them). Thus the first arrow is the zero map so the second arrow in injective (trivial kernel), and so $H_1(S^1)$ is isomorphic to its image in $H_0(U\cap V)$, which is equal to the kernel of the last map, $g_*$. To understand this kernel, we need to remember what this map does.

In general we have inclusions $i:U\cap V\rightarrow U$ and $j:U\cap V\rightarrow V$, inducing homomorphisms in homology. For $\alpha\in H_0(U\cap V)$, then by definition $g_*(\alpha)=(i_*(\alpha),-j_*(\alpha))$. (The $-$ in the second coordinate is to ensure exactness)

More specifically to our case, $U\cap V$ has two components, so $H_0(U\cap V)\cong \mathbb{Z}\oplus\mathbb{Z}$. This is why he says the elements in this group are of the form $ax+by$. Then $g_*(ax+by)=(i_*(ax+by),-j_*(ax+by))$.

Now, finally, we use the path-connectedness of $U$ to say that $i_*(ax+by)=(a+b)z$ where $z$ generates $H_0(U)\cong\mathbb{Z}$ (draw $S^1$, see what's going on geometrically at this step. Recall that $x$ and $y$ are homology classes of points, and as such $i_*(x)$ and $i_*(y)$, sitting in a path-connected space, are homologous.) It then follows from this last equation that $i_*(ax+by)=0$ iff $a=-b$. By an identical argument, ker$j_*$ has the same condition. Then $ax+by\in$ker$g_*$ iff $a=-b$, and so $H_1(S^1)\cong$ker$g_*\cong\mathbb{Z}$

For your second question, as discussed in the comments up to homotopy you should get a wedge of two circles. Consider the Möbius band as a square with opposite sides identified with a half twist. If we take an interior point away from that square, we can retract onto the boundary "frame". Then if we identify opposite sides we get a "$\Theta$ space," homeomorphic to the Greek letter. Then we can just shrink the line in the middle to make an "$8$"

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what a relief! I was feeling sick of not understanding and I did not want to just cram. Thank you very much. The sentence that I take home findly is this one "Recall that x and y are homology classes of points, and as such $i_∗(x)$ and $i_∗(y)$, sitting in a path-connected space, are homologous." – Herband Feb 26 '12 at 19:24
I think I have a solution for my second question. The Mobius band minus one interior point is homotopy equivalent to the wedge sum of two circles(i.e. two circles attached at a point). Am I right? – Herband Feb 26 '12 at 20:03
Oh, I was thinking up-to homeomorphism. I believe you are right! I will edit my solution to include that – you Feb 26 '12 at 22:46