TITLE: Difference between bijection and isomorphism?
QUESTION [55 upvotes]: First, let me admit that I suffer from a fundamental confusion here, and so I will likely say something wrong. No pretenses here, just looking for a good explanation. 
There is a theorem from linear algebra that two vector spaces are isomorphic if and only if they have the same dimension. It is also well-known that two sets have the same cardinality if and only if there exists a bijection between them. Herein lies the issue...
Obviously $|\mathbb{R}| = |\mathbb{R^2}| = \mathfrak c.$ This is often stated as "there are as many points in the plane as there are on a line." Why, then, are $\mathbb{R}$ and $\mathbb{R^2}$ not isomorphic?
It makes intuitive sense that they shouldn't be. After all, I can only "match up" each point in $\mathbb{R}$ with the first coordinate of $\mathbb{R^2}.$ I cannot trust this intuition, however, because it fails when considering the possibility of a bijection $f : \mathbb{N} \rightarrow \mathbb{Q}!$
Even more confusing: As real vector spaces, $\mathbb{C}$ is isomorphic to $\mathbb{R^2}.$ However there is a bijection between $\mathbb{C}$ and $\mathbb{R}$ (just consider the line $\rightarrow$ plane example as above).
If you can explain the error in my thinking, please help!

REPLY [5 votes]: As many others have pointed out already, an isomorphism (in the context of algebra) needs to map between the operations involved, while a bijection doesn't.  A bijection comes as a one-to-one onto map between two unstructured sets.  An isomorphism, on the other hand, maps between two structured sets of the same type or maps between the carriers of two structured sets, or one might say an isomorphism consists of a bijective map between two sets which also satisfies the requisite homomorphic equation(s) for the structures in question.  So, you ask, why aren't R and R^2 isomorphic?  Well, we could have systems (R, +) and (R^2, +, *) which aren't even of the same type and thus it doesn't make any sense to claim that R and R^2 are isomorphic.
As a simple example, consider the set {0, 1} under the minimum operation or ({0, 1}, min) for short, and ({2, 3}, max) where max stands for the maximum operation.  You can (I hope easily) find two bijections between {0, 1} and {2, 3}.  But, there only exists one isomorphism between ({0, 1}, min) and ({2, 3}, max).