Connected sum of surfaces through labels

Question

I have been reading a little bit about the classification of surfaces and we start with the labels, i thought i understood this process but now im trying to show that $P^2 \# T^2$=$P^2\#K^2$=$P^2\#P^2\#P^2$ using labels, manipulating their corresponding labels algebraically. I know that geometrically this is going to be true and it makes sense to me but i really wanted to do this using their labels and manipulating them, so any help is appreciated. Thanks in Advance.

Answer 1

Alright after some time i have a found a way of seeing this using the properties of labels that were a bit confusing to me, im gonna post this here because there might be someone with this doube in the future:

We know that $P^2 \# K^2$ is going to be the label $aa,bcbc^{-1} \rightarrow aac^{-1},cbcbc{-1} \rightarrow ac^{-1}a,bcb \rightarrow a^{-1}ca^{-1},bcb \rightarrow acabcb$

And also we know that

Next we know that $K^2 = P^2 \# P^2$ , geometrically this makes sense since the klein bottle is joining two mobius bands by the border and $P^2$ is a mobius band and a disk attached, but we can see this using labels too since , $abab^{-1} \rightarrow abc^{-1}cab^{-1} \rightarrow bc^{-1}a a{-1}c^{-1}b \rightarrow bccb \rightarrow bbcc$.

Now we know that $P^2 \# K^2$ = $P^2 \# P^2 \# P^2$.

Now we just need to see that the label of $P^2 \# K^2$ is $acabcb$ to see that they are homeomorphic.So we just do some operations on the label:

We start with $ccaba^{-1}b^{-1) \rightarrow ccb^{-1}baba^{-1}b^{-1} \rightarrow ccbaba^{-1} \rightarrow cca^{-1}ababa^{-1} \rightarrow ccabab \rightarrow cacbab \rightarrow cacabb \rightarrow acabbc \rightarrow acabcb$ as we wanted to see. So we have our result proved.