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

On a introductory Topology course, I learned how to make some connected sums of spaces informally. For example, in the figure below, we have a draft for $\mathbb{R}P^2\ \# \ \mathbb{R}P^2$, where $\mathbb{R}P^2$ is the real projective plane.

First, we take two copies of $\mathbb{R}P^2$, as shown in $(i)$. Then, we ''glue'' them making ''holes'', one of which contains $A$. The sequences of arrows $p$-$q$ and $r$-$s$ can be simplified, and the holes create a new arrow $c$, given us $(ii)$. I regarded the holes in the copies as pieces of the squares containing the lower left vertices. The border of these pieces are the arrows $c$. Then, after some transformations displayed in $(iii)$, we get $K$, the Klein bottle.

Then, I tried the same method to prove that $\mathbb{R}P^2\ \# \ K$ and $\mathbb{R}P^2\ \# \ \mathbb{T}^2$ are homeomorphic, but that becomes more complex since the possibilities of cuts and transformations increase. I'd like to know if someone could give me a suggestion. I could find the homology groups of theses spaces, but I think it doesn't suffice. Thank you!

• Well, I found out that the page math.stackexchange.com/questions/358724/… has the same question. But I'm not suppose to know Euler characteristic on my course and I couldn't understand the hint of the picture presented there.
– rgm
Feb 2, 2017 at 12:06

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 $$\mathbb RP^2\ \# \ \mathbb K$$.