Past papers at my University frequently ask for a geometric argument that proves that some shape with side identifications is homeomorphic to another shape, e.g. a torus. Quite frequently there are some strange arguments for this; I'm fine with connecting two shapes together, turning two lines into one if they are next to eachother and in the same direction, reflecting and rotating; but sometimes other things are done, like removing a shape and gluing it on at a different place; is there any foolproof way to know what to do and when, or are there any tips and tricks for deciding what to do that I might be missing?
An algebraic argument would also be fine, though I find it a little harder to think about in this way.
Sorry for the loosely defined question, and thanks for any help you can give!
"removing a shape and gluing it on at a different place" is weird, but
"turning two lines (edges?) into one if they are next to each other" is OK.
That second operation is part 2 of the first one, so it must be part 1 of the first operation that bugs you.
Here's a lemma for you:
If $R$ is an equivalence relation on $X$, and $S$ an equivalence relation on $Y$, and there's a map $f$ from $X$ to $Y$ with the property that $x R x'$ if and only if $f(x) S f(x')$, then there's a bijection, $\hat(f)$, from $X / R$ to $Y / S$ with the property that $\hat(f)([x])= [f(x)]$ for every element $x \in X$, where $[x]$ denotes the $R$-equivalence class of $x$, and $[f(x)]$ denotes the $S$-equivalence class of $f(x)$.
You can probably prove this lemma.
Now consider a polygon $X$, and slice it into two pieces; call the (disjoint) union of those two pieces $Y$. Let $R$ be the trivial equivalence relation on $X$ (i.e., $x R x'$ iff $x = x'$), and let $S$ be the equivalence relation $Y$ consisting of the trivial relation, together with the relation that says that points in the two "pieces" that lie along the previously-shared edge are "equivalent".
Then the lemma says that $X/R = X$ is in bijection with $Y/S$. In other words, the "cut apart" thing is, with the gluing rule, the same as the original.
This same argument can be extended to handle continuity/topology, with exactly the same consequences: it's OK to cut and paste.
As for "is there a standard rule for how to do these operations?", the answer (for surfaces without boundary) is "yes". If your surface is given to you as several polgyons with edges identified, first glue up edge-pairs until you have only a single polygon for each surface.
Now the classification theorem for surfaces actually consists of an algorithm for sequentially altering such a polygon-description of a surface into a standard one, after which comparison is easy. You apply this algorithm to both your surfaces, and if they come out the same, then they were homeomorphic; if they don't, then they weren't.
I'm not going to repeat the algorithm here -- it's a long chapter in most algebraic topology books -- but it's worth noting that surfaces are incredibly special in this regard: for 3-manifolds (or higher dimensions), there's no known such algorithmic way of testing for homeomorphism.
[Books I like for classification of surfaces: Massey, intro to algebraic topology (or something like that); Seifert and Threllfall, a Textbook of Topology. Very old-fashined, but wonderful.]