Let $M$ be a smooth manifold and suppose that $\Sigma \subset M \times[0,1]$ is a surface such that $\partial \Sigma \subset \partial (M \times [0,1])= M \times \{0,1\}$. Denote by $\pi$ the projection onto the first factor for $M \times[0,1]$. Since $\Sigma$ admits a triangulation whose boundary is a triangulation of $\partial \Sigma$, it follows that $[\pi(\partial \Sigma)] \in H_1(M;\mathbb{Z})$ is zero.
This seems like a perfectly reasonable argument to me. Nevertheless, when I started to work through some examples, I found that this is readily demonstrated to be false even in the simplest case of $M$ being the torus considered as a quotient of $[0,1]\times [0,1]$ in the standard way. Start off with the loop $\{1/2\} \times [0,1]$ and pull the component $\{1/2\} \times (2/5,3/5)$ around longitudinally until it (almost) touches itself at the point $(1/2,4/5)$, then perform surgery which corresponds to a saddle cobordism (by cutting out two arcs and gluing them back to the 4 resulting boundary points, but the other way around). The resultant curve is a meridian plus two longitudes at the level of homology. Where am I going wrong in my reasoning?? Any and all insights are welcome.