Prove that $\mathbb{R}P^2\ \# \ K$ and $\mathbb{R}P^2\ \# \ \mathbb{T}^2$ are homeomorphic.

Solution 1:

In these notes: https://people.maths.ox.ac.uk/hitchin/hitchinnotes/Geometry_of_surfaces/Chapter_1_Topology.pdf, Nigel Hitchin produces the following solution:

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where the first figure represents $\mathbb RP^2\ \# \ \mathbb T^2$. Pasting the last two quadrilaterals by the sides with the right triangles, we obtain $\mathbb RP^2\ \# \ \mathbb K$.

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