What's the difference between a bijection and an isomorphism? I don't understand what the difference is between a bijection and an isomorphism.  They seem to both just be a invertible mapping.
Is the set of all bijections a subset of isomorphisms?  Or vice versa?  What is the difference and can you show me an example of one that is not the other?
Thanks.
 A: The answer is "vice versa." An isomorphism is a structure-preserving bijection. The specific meaning of "structure" will vary, depending on the context.
A: If you are talking just about sets, with no structure, the two concepts are identical. Usually the term "isomorphism" is used when there is some additional structure on the set. For example, if the sets are groups, then an isomorphism is a bijection that preserves the operation in the groups: $\varphi(ab) = \varphi(a)\varphi(b)$. As another example, if the sets are vector spaces, then an isomorphism is a bijection that preserves vector addition and scalar multiplication.
A: As Cameron says, an isomorphism is a structure-preserving bijection, like a group isomorphism, which is a bijection that preserves the group structure. Or ring isomorphisms. Or algebra isomorphisms. Or module isomorphisms. Or ...
Homeomorphisms are the mappings that preserve the topological properties (more precisely, they are isomorphisms in the category of topological spaces). Wikipedia gives a counterexample to the statement that every bijection is an isomorphism:

Consider for instance the function $f: [0, 2π) → S^1$ (the unit circle in $\mathbb{R}^2$) defined by $f(φ) = (\cos(φ), \sin(φ))$. This function is bijective and continuous, but not a homeomorphism ($S^1$ is compact but $[0, 2π)$ is not). The function $f^{−1}$ is not continuous at the point $(1, 0)$, because although $f^{−1}$ maps $(1, 0)$ to $0$, any neighbourhood of this point also includes points that the function maps close to $1$, but the points it maps to numbers in between lie outside the neighbourhood.

