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Problem 2.2.30 in Hatcher involves the homomorphisms $H_2(S^1\times S^1) \cong \mathbb{Z} \rightarrow H_2(S^1\times S^1) \cong \mathbb{Z}$ induced by

The map $S^1 \times S^1 \rightarrow S^1 \times S^1$ that is the identity on one factor and a reflection on the other.

and

The map $S^1 \times S^1 \rightarrow S^1 \times S^1$ that interchanges the two factors and then reflects one of the factors.

I strongly suspect that the induced homomorphisms are $-\mathbf{1}$ and $\mathbf{1}$, respectively. How would I prove this?

user109360
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    Are you familiar with the Künneth formula? – Ayman Hourieh Sep 07 '14 at 21:51
  • I want to avoid it because Hatcher does not mention the formula until Chapter 3. – user109360 Sep 07 '14 at 21:54
  • Would simplicial homology work? It seems concrete enough that you could directly compute it. – DCT Sep 07 '14 at 22:07
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    Simplicial homology would work, and so would cellular homology. Find the generator(s) of each homology group and see what $f$ maps them to. – Ayman Hourieh Sep 07 '14 at 22:15
  • I tried to use the $\Delta$-complex structure on page 102 of Hatcher in which a square is divided into two triangles across a diagonal. Then $H_2$ is generated by $U-L$, where $U$ and $L$ are the upper and lower triangles. Switching the factors interchanges $L$ and $U$ because it reflects the square across the diagonal. I think that the map reflecting one factor corresponds to either a horizontal or vertical reflection, but both of these reflections result in a square with the diagonal going the other way. – user109360 Sep 07 '14 at 22:37
  • Yes, that is annoying, but you can take a $\Delta$-complex structure that is preserved under switching factors and reflection. For example, you can draw both diagonals, so you have 4 triangles instead of 2. – DCT Sep 08 '14 at 00:54

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