In Hatcher's Algebraic Topology, Chapter 1, page 51, he describes two presentations of the Klein bottle:
The first one is the usual one, a square with opposite sides identified via the word $aba^{−1}b$, then Hatcher says that if one cuts the square along a diagonal and reassembles the resulting two triangles as shown in the figure, one obtains the other representation as a square with sides identified via the word $a^2c^2$.
Question: Does this process, which involves cutting and gluing, automatically imply the two presentations are equivalent?
Just to summarize the comments:
The Klein bottle are two triangles with edges identified as follows:
We recognize the Klein bottle solely by the cycles on the boundaries of the two triangles, which are $cab$ and $cb^{-1}a$. It doesn't matter how we arrange the two triangles, they can be put separately as above, or two $a$'s putting together, or two $b$'s putting together, or two $c$'s putting together.