I have a problem of classification of topological surfaces.
Let S be the surface given by the sequence of identifications A B C B E where A,B,C,E are sequences of edges. Let $\beta$ be a letter which doesn't appear in the previous sequences.
I have to demonstrate that S is homeomorphic to the surface obtained by the sequence A $\eta$ C $\eta$ E, where we substitute the entire sequence B with the only edge $\eta$.
I think I have to use the classification theorem, but I don't understand how a sequence of possible different edges (ex. B = $\alpha\gamma\delta$ ) can be reduced to a single edge, given that in such a sequence no eliminations of edges are possible.
Where am I proceeding wrong?
You don't need the classification of surfaces.
Let's take the example you suggest, where $B = \alpha \gamma \delta$ is a sequence of three edges. There are two occurrences of $B$; let those two occurences be denoted $B_1 = \alpha_1 \gamma_1 \delta_1$ and $B_2 = \alpha_2 \gamma_2 \delta_2$. Consider the two points $p_1 = \alpha_1 \cap \gamma_1$ and $p_2 = \alpha_2 \cap \gamma_2$. The two vertices $p_1,p_2$ are identified with each other when the $B_1$ and $B_2$ sequences are glued. Furthermore, $p_1,p_2$ form a complete vertex cycle, and therefore no other point is identified with them. We can therefore rewrite $B_1$ and $B_2$ in the following form: $$B_1 = \underbrace{\zeta_1}_{\alpha_1 \gamma_1} \delta_1 $$ $$B_2 = \underbrace{\zeta_2}_{\alpha_2 \gamma_2} \delta_2 $$ In other words, we have removed $p_1$ and $p_2$ from the list of vertices, conglomerating the sequence $\alpha_i\gamma_i$ into a single edge $\zeta_i$, for each of $i=1,2$. So the expression of $B_i$ as a sequence of three edges has been converted into an expression as a sequence of only two edges.
The same argument is the basis for an induction: if $B$ is expressed as a sequence of $k$ edges then, by conglomerating, I can rewrite $B$ as a sequence of $k-1$ edges. By induction, I can get $B$ down to a single edge.