Bicontinuous part of Homeomorphism definition By definition, a function $f:X\rightarrow Y$ is said to be a homeomorphism provided that $f$ is bijective and bicontinuous. I am aware of the intuition that a homeomorphism deforms one space into another via stretching and bending, but not tearing or gluing. I'm wondering what the motivation is for the bicontinuous condition? It is clear that at least one direction should be continuous, but why both? Is there some geometric intuition for why we require $f$ to be bicontinuous, or is it simply due to the fact that it works out nicely and allows a homeomorphism to be a morphism of categories that has a bunch of nice properties? 
 A: In a very rough description, the continuity of $f$ excludes transformations which "tear" or "punch holes in" a space.  The continuity of $f^{-1}$ excludes transformations which "glue".
If $f$ is continuous, and $X$ is connected, then $f(X)$ is also connected.  So, $X$ will not be "torn" into more than one "piece".
As for "gluing", you are right that the function needs to be bijective, but we also need $f^{-1}$ to be continuous.  Consider $f:[-2,1]\cup (1,2]\to [0,2]$ given by
$$f(x)=\begin{cases}x+2, & x\in [-2,-1]\\x, & x\in(1,2].\end{cases}$$
It's easy to see that $f$ is continuous and bijective. However, we see that the two "pieces" $[-2,-1]$ and $(1,2]$ are "glued" together to get $[0,2]$.
A: Not really a geometric intuition but a set-theoretic one: a bijection $f \colon A \to B$ can be continuous without its inverse being continuous. For example, for any nonempty set $X$:
$$
x \mapsto x \colon (X, discreteTopology) \to (X, indiscreteTopology)
$$
is a continuous bijection, 
but if $|X| > 1$, $f^{-1}$ is not continuous.
Th analogous trivial example holds when the topology of the domain is strictly finer than that of the codomain.
A: The idea of a homeomorphism $f:A\to B$ is that topologically $A$ and $B$ are the "same" space. That is, the function $f$ induces a bijection $f^*$ from the topology $T_A$  on $A$ to the topology to the topology $T_B$ on $B$, where $f^*(t)=\{f(a) :a \in t\}$. There are countless examples of a continuous bijection $g$ not being a homeomorphism, as the image of an open set of $A$ may to fail to be open in $B$ . A simple example is $A=\{0\}\cup (1,2]$ and $B=[1,2]$ with $g(0)=1$ and $g(x)=x$ for $x\in (1,2]$. There are also many subtle or complicated examples.
