# Explicit construction of map from $T^2\rightarrow S^2$ with lowest non-zero degree of map.

This question is related to my another question:Homotopy class of map from torus to sphere?

I want to construct all different homotopy class of map from $$T^n \rightarrow S^n$$. From above question, the homotopy class is isomorphic to $$m\mathbb{Z}\cong \mathbb{Z}$$ with $$m$$ the lowest non-zero degree of map.

My question:

1. What's the $$m$$ for the case $$T^n\rightarrow S^n$$?

2. How to explicitly construct the map with lowest non-zero degree of map? Then I can combine the $$k$$ covers of $$S^n$$ to $$S^n$$ to get all different homotopy class of $$T^n$$ to $$S^n$$.

• A closed, connected, orientable manifold always admits a degree one map to a sphere of the same dimension. See this question for example. – Michael Albanese Nov 7 '18 at 12:21