Homology of an edge with the endpoints identified I was studying Mayer Vietoris Long Exact Sequence. To get my hands dirty I wanted to try some examples: let $L$ be two points and $K$ be two points connected with an edge, in euclidean space. My aim is to compute homology of $K/L$ via this technique.
More precisely, let $L = \{\{u\},\{v\}\}$ and $K = \{\{u\},\{v\}, \{u,v\}\}$ be two simplical complexes.
Now my understanding of $0^{th}$ and $1^{st}$ homology groups is that, they represent connected components and tunnels, respectively. In particular, $K/L$ is a loop and so it has one connected component and one tunnel and hence I would expect the relative homology to be $H_0(K,L) = \mathbb{F} = H_1(K,L)$ and $H_i(K,L) = 0$ for $i \geq 2$.
I will now attempt to do the computations to see if this is true indeed: We have the following short exact sequence $0 \to C_d(L) \to C_d(K) \to C_d(K,L) \to 0$ for each $d \geq 0$. This gives rise to the long exact sequence  $0 \to H_1(K,L) \to H_0(L) \to H_0(K) \to H_0(K,L) \to 0$.
Here we have $H_0(L) = \mathbb{F}^2$, $H_0(K) = \mathbb{F}$ and $H_0(L) \to H_0(K)$ is a rank $1$ map. Which gives $H_1(K,L) = \mathbb{F}$ but $H_0(K,L) = 0$!
First I doubted my calculations but then we do have $C_0(K,L) = C_0(K)/C_0(L) = 0$ and $C_1(K,L) = C_1(K)/C_1(L) = \mathbb{F}$, so my computation seems correct, right?
How is $K/L$ different from, say the graph $K_3$ or $S^1$ geometrically? This has left me questioning my intuition and I would really appreciate if someone could help me understand this. Also, I would like to ask what is it exactly that the zeroth homology group represent here?
 A: It seems that your confusion is coming from mixing up the homology of $K/L$ and the homology of the pair $(K,L)$; they're not quite the same.
First, your computation is correct: $H_0(K,L) \cong 0$, $H_1(K,L) \cong \mathbb{F}$, and all other homology groups are zero.
When you earlier try to intuit the homology groups, you actually correctly guessed the homology groups of $K/L$ (which is indeed homeomorphic to $S^1$), with $H_0$ and $H_1$ rank $1$ and all other homology groups zero. So there is a difference between $H_*(K,L)$ and $H_*(K/L)$.
Relative homology of a pair $(X,A)$ is capturing "homology of $X$ modulo $A$." That is, an element of $H_k(X,A)$ is a $k$-chain in $X$ whose boundary lies in $A$, and is trivial if it is the boundary of a $(k+1)$-chain in $X$ (modulo the subgroup $C_k(A)$). Thus $H_0(X,A)$ is free of rank equal to the number of path components of $X$ that do not intersect $A$.
Relative homology of a pair $(X,A)$ is almost the same as the homology of the quotient $X/A$. If the pair $(X,A)$ satisfies the hypotheses that $A$ is closed and nonempty and there exists a neighborhood $U$ of $A$ in $X$ such that $U$ deformation retracts to $A$ (this is referred to in various places as a "good pair" or an "NDR pair" (stands for 'neighborhood deformation retract'), and is also equivalent to the requirement that the inclusion $A \to X$ is a cofibration), then excision of singular homology is equivalent to the statement that $H_k(X,A) \cong \tilde{H}_k(X/A)$. An important thing to note is that when the pair $(X,A)$ consists of a (simplicial, CW) complex and a subcomplex, the pair always satisfies this hypothesis.
Since reduced homology coincides with absolute homology above degree zero, we have $H_1(K,L) \cong H_1(K/L)$. In view of this theorem, the difference between $H_0(K,L)$ and $H_0(K/L)$ is precisely one factor of $\mathbb{F}$ since $H_0(K/L) \cong \tilde{H}_0(K/L) \oplus \mathbb{F}$ and $K/L$ is path-connected so $H_0(K/L) \cong \mathbb{F}$, giving $0 = \tilde{H}_0(K/L) \cong H_0(K,L)$.
References for these facts can probably be found in most books on singular homology or introductory algebraic topology books. I only know the specific locations of references from Hatcher's book, so: discussion of relative homology is in section 2.1 (pages 115-116), the theorem establishing equivalence of relative homology of an NDR pair and the reduced homology of the quotient is proposition 2.22, and the fact that a CW pair $(X,A)$ with $A$ nonempty is always an NDR pair follows from proposition A.5 or 0.16.
