Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 20H (Algebraic Topology)

Subdivide the torus so $p,q$ are vertices and add an edge path between them; identifying its endpoints yields a finite simplicial model, so $X$ is triangulable.

Identifying two points in a path-connected space is homotopy equivalent to adjoining a circle, so $X\simeq T^2\vee S^1$. Hence $H_0=\mathbb{Z}$, $H_1={\mathbb{Z}}^3$, $H_2=\mathbb{Z}$, and all higher homology vanishes.

No. Collapse the one-skeleton of the torus summand to obtain a map $X\to S^2$ inducing degree one on $H_2$. A null-homotopic map induces zero on reduced homology.