On page 449 of Lee's Introduction to Smooth Manifolds (2nd Edition), the Mayer-Vietoris Theorem is given:
Let $M$ be a smooth manifold. Let $U$ and $V$ be open in $M$ such that $U\cup V=M$. Then for each $p$ there is a linear map $\delta$ such that the following sequence is exact:
$$\dots \xrightarrow{\delta} H^p(M) \xrightarrow{k^*\oplus l^*} H^p(U)\oplus H^p(V) \xrightarrow{i^*-j^*} H^p(U\cap V) \xrightarrow{\delta} H^{p+1}(M) \xrightarrow{k^*\oplus l^*} \dots$$
My question is: Suppose $p=0$. Is the left most arrow in the diagram above mapping $0$ into $H^0(M)$?
As has been mentioned in the comments, you are correct. The beginning of the Mayer-Vietoris sequence is
$$0 \to H^0(M) \xrightarrow{k^*\oplus l^*} H^0(U)\oplus H^0(V) \xrightarrow{i^*-j^*} H^0(U\cap V) \xrightarrow{\delta} H^1(M) \to \dots$$
and the end is
$$\dots \to H^{n-1}(U\cap V) \xrightarrow{\delta} H^n(M) \xrightarrow{k^*\oplus l^*} H^n(U)\oplus H^n(V) \xrightarrow{i^*-j^*} H^n(U\cap V) \to 0.$$