Let $\mathcal{H}_*$ be a ordinary homology theory, that is, a homology theory that satisfies the Eilenburg-Steenrod Axioms. I wanted to show that $\mathcal{H}_1(S^1)\cong \mathcal{H}_0(S^1)\cong R$, where $H_0(*)\cong R$, by using the Mayer-Vietoris sequence.
The standard way to do this is to use open cover $U:=S^1\setminus\{N\}, V:=S^1\setminus\{S\}$ where $N,S$ are the north and south poles respectively. Then, we get a sequence
$$0\rightarrow \mathcal{H}_1(S^1)\rightarrow \mathcal{H}_0(U\cap V)\xrightarrow{\mathcal{H}_0(i_U)\oplus \mathcal{H}_0(i_V)} \mathcal{H}_0(U)\oplus \mathcal{H}_0(V)\rightarrow \mathcal{H}_0(S^1)\rightarrow 0,$$
where $i_U$ and $i_V$ are the inclusions of $U\cap V$ in $U$ and $V$ respectively.
Now, we have that $U\cap V$ is a homotopy equivalent to the disjoint union of two points and both $U$ and $V$ are homotopy equivalent to a point and so our sequence becomes $$0\rightarrow \mathcal{H}_1(S^1)\rightarrow R\oplus R\xrightarrow{\lambda} R\oplus R\rightarrow \mathcal{H}_0(S^1)\rightarrow 0.$$
What I'm having trouble with is understandng what the map lambda will look like; my professor said it would be given by the matrix $\begin{bmatrix}1 & 1\\ 1& 1\end{bmatrix}$, in which case, we would have that $\mathcal{H}_1(S^1)\cong \mathcal{H}_0(S^1)\cong R$ as expected,but I can't see why this is true; in particular, why would the map look like $$\lambda:(a,b)\mapsto (a+b,a+b)?$$ Furthermore, in general, how does one track the changing og the inclusion map under any ordinary homology theory?
$\require{AMScd}\newcommand{\H}{\mathcal{H}}$This is not so bad. We just need to understand how the homotopy equivalences $U\simeq\{\text{pt}\}$ etc. affect the inclusions.
$$\begin{CD}R\oplus R@>\begin{pmatrix}1&1\\1&1\end{pmatrix}>>R\oplus R\\@V\cong VV@VV\cong V\\\H_0(\{W\})\oplus\H_0(\{E\})@>\begin{pmatrix}\H_0(i_{W,V})&\H_0(i_{E,V})\\\H_0(i_{W,U})&\H_0(i_{E,U})\end{pmatrix}>>\H_0(V)\oplus\H_0(U)\\@V\H_0(i_W)+\H_0(i_E)V\cong V@VV=V\\\H_0(U\cap V)@>>i_0\H_0(i_V)+i_1\H_0(i_U)>\H_0(V)\oplus\H_0(U)\\\end{CD}$$Commutes simply by functoriality, where I have taken the West and East 'poles' of the circle. Every element of the matrix in the middle row is $\H_0(f)$ for some $f$ homotopic to the identity, justifying the replacement by "$1$". Inverting isomorphisms where necessary we get the claim.