Why $H_1(SX)\cong \widetilde{H_0}(X)$?
Suppose that $X$ is a topological space and $SX$ is the suspension of this space, so if we take as open $U=SX-\{\overline{(x,1)}\}$ and $V=SX-\{\overline{(x,0)}\}$ for any $x\in X$, we see that $U,V$ are contractible and that $U\cap V \simeq X$, then the sequence of Mayer-Vietoris is the following exact sequence
$$...\to H_n(U\cap V)\to H_n(U)\oplus H_n(V)\to H_n(SX)\to \\ H_{n-1}(U\cap V)\to H_{n-1}(U)\oplus H_{n-1}(V)\to...$$
So if $n>1$ then it is clear that $H_n(SX)\cong H_{n-1}(X)$, but why would $H_1(SX)\cong \widetilde{H_0}(X)$, could someone explain please? thank you.
The question was answered in cjackal's comment and this community wiki intends to clear the question from the unanswered queue.
The Mayer-Vietoris has a variant for unreduced homology and a variant for reduced homology. See for example the section "Mayer–Vietoris Sequences" in Allen Hatcher, Algebraic topology. The reduced homology groups $\tilde{H}_n(Z)$ agree with the unreduced homology groups $H_n(Z)$ for $ n > 0$, whereas $\tilde{H}_0(Z)$ is the kernel of the map $H_0(Z) \to \mathbb Z$ induced by $\epsilon : C_0(Z) \to \mathbb Z, \epsilon(\sum n_i \sigma_i) = \sum n_i$, which implies $H_0(Z) \approx \tilde{H}_0(Z) \oplus \mathbb Z$. See Hatcher (text after Proposition 2.8).
Now use the reduced Mayer-Vietoris sequence and observe that $\tilde{H}_0(Z) \approx 0$ for contractible $Z$.