Computing singular homology of a cylinder with a bottom, using Mayer Vietoris Sequence I am trying to practice using MVS on an easy example, a cylinder with a bottom. Explicitly, something like $S^1 \times [0,1]$ with a copy of $D^2$ glued at one end.
Call this object $X$. Then I propose splitting $X$ into $A\cup B$ where $A$ is an open cylinder, $B$ is a disk, and therefore $A\cap B$ is homotopic to $S^1$. Then $H_n (A) = \mathbb{Z}$ if $n=0,1$, and $0$ otherwise. $H_n (B)= \mathbb{Z}$ for $n=0$ ($0$ otherwise), and $H_n (A\cap B) = H_n (A)$ since the open cylinder is homotopic to a circle.
Now I am trying to compute $H_1 (X)$ but having trouble understanding explicitly what the maps will be. I already know that $H_1 (X) \cong \pi_1 (X) \cong 1$, but this is just for practice finding maps.
$$H_1 (A\cap B) \xrightarrow{f} H_1 (A) \oplus H_1 (B) \xrightarrow{g} H_1 (X) \xrightarrow{h} H_0 (A \cap B) \xrightarrow{i} H_0 (A) \oplus H_0 (B)$$
I know all the terms now except $H_1 (X)$, so I fill the sequence in to get,
$$\mathbb{Z} \xrightarrow{f} \mathbb{Z} \oplus 0 \xrightarrow{g} H_1 (X) \xrightarrow{h} \mathbb{Z} \xrightarrow{i} \mathbb{Z} \oplus \mathbb{Z}$$
I think the map $i$ will represent the inclusion of $A\cap B$ into $A$ and into $B$, respectively, but what does this actually mean in this case?
In other posts on similar problems I've seen that it's often helpful to convert the objects into simplicial complexes?
I've re-read this section of Hatcher several times, but can't seem to grasp how to accomplish finding maps in MVS in general, or at least what precisely the goal is when computing them. I can understand the specific examples done in the book (e.g., torus and Klein-bottle) but immediately get lost when trying to apply it to a different space.
 A: Note first that since $f$ is an isomorphism, then $g=0,$ so to show that $H_1(X) = 0$ you'd only like to have that $i$ in injective, so that $h = 0$ and $H_1$ is sandwiched between two zeros.
But anyway, $H_0$ counts connected components of spaces, and on maps it corresponds to connected component inclusion, for example $H_0(\bullet \> \bullet) \to H_0(\bullet \> \bullet \> \bullet)$ is a map $\mathbb{Z}^2 \to \mathbb{Z}^3$ that sends generators corresponding to both of two points to generators corresponding to these same two points in the three-point space. In your case, everyone has one connected component, so both $H_0 (A\cap B) \xrightarrow{} H_0 (A)$ and $H_0 (A\cap B) \xrightarrow{} H_0 (B)$ are isomorphisms. Then $i$ is a diagonal.
As to how to find maps of homology in general: well, you look how (algebraic) generators like $1 \in \mathbb{Z}$ are represented by geometric "pieces" and trace where those pieces are sent. In general, the two most common ways of viewing what "pieces" are come from simplicial or cellular homology. In the case of $H_0,$ however, it is easy because every generator is any point in a connected component. So you take, say, a point in  $A \cap B.$ It generates $H_0.$ Its image under $A \cap B \to A$ is a point in $A$. It generates $H_0$ there. You conclude that the map of zero homology groups is an isomorphism.
