1. **Problem statement:** Show that the intersection of two circles is a retract of the torus minus a point. 2. **Understanding the problem:** The torus $T^2$ can be viewed as $S^1 \times S^1$, the product of two circles. Removing a point from the torus, $T^2 \setminus \{p\}$, is homotopy equivalent to a wedge sum of two circles $S^1 \vee S^1$. 3. **Key fact:** The intersection of two circles in the torus corresponds to the wedge sum $S^1 \vee S^1$, which is a subspace of $T^2 \setminus \{p\}$. 4. **Goal:** Construct a retraction map $r: T^2 \setminus \{p\} \to S^1 \vee S^1$ such that $r$ restricted to $S^1 \vee S^1$ is the identity. 5. **Construction:** Since $T^2 \setminus \{p\}$ deformation retracts onto $S^1 \vee S^1$, the intersection of the two circles is a retract of $T^2 \setminus \{p\}$. 6. **Summary:** The intersection of two circles in the torus is homeomorphic to $S^1 \vee S^1$, and since $T^2 \setminus \{p\}$ deformation retracts onto this wedge sum, the intersection is a retract of the torus minus a point. **Final answer:** The intersection of two circles is a retract of the torus minus a point because $T^2 \setminus \{p\}$ deformation retracts onto $S^1 \vee S^1$, which is homeomorphic to the intersection of the two circles.