Let $X = S^1 \times I$. I know that $H_1(X) = \mathbb{Z}$, which implies that $H_1(X)$ has 1 generator. It is intuitive to see that the generator of $H_1(X)$ is the circle that wraps around the cylinder. However, when I look at the definition of $C_1(X)$, it is the free module over the set of all continuous functions $\sigma: \triangle^1 \rightarrow X$. There are infinitely many $\sigma$,i.e,circle that wraps around $X$. So somehow, passing from $C_1(X)$ into $H_1(X)$, all of these circles become 1 equivalent class. Can anyone show me how this is true from the definition of chain complexes and homology.
Generator of first homology group of cylinder
algebraic-topologyhomology-cohomology
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I think people should ignore all this nonsense about counting holes and just look at what actually happens in more examples.
In particular, your intuition about top homology having something to do with enclosing volumes is not quite correct. I interpret this to mean that you have in mind a manifold which is the boundary of another manifold (the same way that the sphere $S^n$ is the boundary of the disk $D^{n+1}$), and it's not true that a manifold has to be a boundary in order to have nonvanishing top homology. The simplest closed counterexample is $4$-dimensional: the complex projective plane $\mathbb{CP}^2$ is a $4$-manifold with $H_4 = \mathbb{Z}$, but it's known not to be the boundary of a $5$-manifold.
Whether top homology vanishes or not instead has to do with orientability.
It's possible to directly visualize the case of $\mathbb{RP}^2$, so let's do it. In this case $\pi_1 \cong H_1 \cong \mathbb{Z}_2$, so the goal is to visualize why there's some loop which isn't trivial but such that twice that loop is trivial. Visualize $\mathbb{RP}^2$ as a disk $D^2$, but where antipodal points on the boundary have been identified. We'll consider loops starting and ending at the origin.
I claim that a representative of a generator of $\pi_1 \cong H_1$ is given by the loop which starts at the origin, goes up to the boundary, gets identified with the opposite point, and goes up back to the origin. Try nudging this loop around for a bit so you believe that it's not nullhomotopic: the point is that you can't nudge it away from the boundary because the two (antipodal, hence identified) points it's intersecting the boundary at can never annihilate.
Now we want to visualize why twice this loop is nullhomotopic. It will be convenient to nudge the loop so that it hits the boundary at four points, which come in two antipodal pairs $A, A', B, B'$, so that the loop hits them in that order before returning to the origin. At this point it would be helpful to draw a diagram if you haven't already; the disk, and the two loops in it, should look a bit like a tennis ball in profile. Now, nudge the loop so that $A, B'$ get closer together, and hence, since they're constrained to be antipodes, $A', B$ also get closer together. Eventually you'll have nudged them enough that you'll see that you can finally pull the curve away from the boundary: as you do so, $A', B$ annihilate each other, and then $A, B'$ annihilate each other.
A similar visualization works for the Klein bottle, visualized as a square with its sides identified appropriately.
The Mayer-Vietoris sequence is defined as the long exact sequence
$$H_*(U \cap V) \to H_*(U) \oplus H_*(V) \to H_*(X)$$
where the first map is defined by $(i_*, j_*)$ and the second map is defined by $k_* - l_*$ where $i, j$ are the inclusion maps of $U \cap V$ inside $U$ and $V$ and $k, l$ are the inclusions of $U$ and $V$ inside $X$.
If $X$ is the Klein bottle, $\{U, V\}$ cover of it by the two mobius strips, then $U \cap V$ is homotopy equivalent to the boundary of the two strips. I have to thus compute the maps $i_*, j_*$ induced on homology where $i, j$ are inclusions of the boundary map of the mobius strips into the Klein bottle.
That's precisely where we need the wrapping description. If $M$ is a moebius strip then the inclusion map $\partial M \hookrightarrow M$ could be understood as follows: deformation retract $M$ to it's core circle, which generates $H_1(M)$. Under this deformation retract $\partial M$ maps to twice the core circle. So $i_* : H_1(\partial M) \to H_1(M)$ is precisely the multiplication by $2$ map.
Thus, $i_*$ is multiplication by $2$ whereas $j_*$ is multiplication by $-2$ due to orientation issues. Tat tells you $\alpha$ sends $1$ to $(-2, 2)$ once you identify $H_1(U \cap V)$ with $\Bbb Z$ and $H_1(U) \oplus H_1(V)$ with $\Bbb Z \oplus \Bbb Z$.
Best Answer
Observe that the cylinder $X= S^1\times [0,1]$ is homotopy equivalent to $S^1\times\{0\} = S^1$, therefore $$H_1(S^1\times[0,1]) = H_1(S^1) = \mathbb{Z}$$
Now if you want to justify $H_1(S^1\times[0,1]) = \mathbb{Z}$ purely from the properties of the singular chain complex $(C_\bullet(X), \partial_\bullet)$, it's much more complicated, that's because computing singular homology groups directly from the chain complex is complicated.
Fortunately, we can equip the cylinder with a $\Delta$-complex structure. The simplicial homology $H_*^\Delta(X)$ which we get from the simplicial chain complex $(C_\bullet^\Delta(X), \partial_\bullet^\Delta)$ computes up to isomorphism singular homology $H_*(X)$. That means, simplicial homology and singular homology are isomorphic homology theories $$H_*^\Delta(X) \cong H_*(X).$$
Using the $\Delta$-complex structure of the cylinder allows you to directly compute the homology groups $H_n(X)$ from the respective chain groups $C_n(X)$.
The $\delta$-complex structure of the cylinder is pictured below
$\hskip2in$
Your chain groups are therefore $$C_0(X) = \langle v,w\rangle,\quad C_1(X) = \langle a,b,c,d\rangle,\quad C_2(X) = \langle A,B\rangle$$
the boundary operator $\partial_n$ is defined analougously as for the singular chain complex, therefore you can evaluate $\partial_n$ on the given generators in each degree and compute your homology groups via $$H_1(X) = \ker \partial_1/{\operatorname{im} \partial_{2}}.$$