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’tbe. 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!

**Answer**

To elaborate a bit on Tobias answer. The notion of isomorphism depends on which structure (category actually) you are studying.

Edit: Pete L. Clark pointed out that I was too sloppy with my original answer.

The idea of an isomorphism is that isomorphisms preserve all structure that one is studying. This means that if X,Y objects in some category, then there exists morphisms f:X\rightarrow Y, g:Y\rightarrow X such that f\circ g is the identity on Y, and g\circ f is the identity on X.

To be a bit more explicit, if X and Y are sets, and there is a bijective function X\rightarrow Y, then we can construct the inverse f^{-1}:Y\rightarrow X. This inverse function is defined by f^{-1}(y)=x iff f(x)=y. We have that f\circ f^{-1}=id_Y and f^{-1}\circ f=id_X.

But if we are talking of vector spaces, we demand more. We want two vector spaces to be isomorphic iff we can realize the above situation by linear maps. This is not always possible, even though there exists a bijection (you cannot construct a invertible linear map \mathbb{R}\rightarrow \mathbb{R}^2). In the linear case; if a function is invertible and linear, its inverse is also linear.

In general however, it need not be the case that the inverse function of some structure preserving map preserves the structure. Pete pointed out that the function x\mapsto x^3 is an invertible function. It is also differentiable. but its inverse is not differentiable in zero. Thus x\mapsto x^3 is not an isomorphism in the category of differentiable manifolds and differentiable maps.

I would like to conclude with the following. We cannot blatantly say that two things are isomorphic. It depends on the context. The isomorphism is always in a category. In the category of sets, isomorphisms are bijections, in the category of vector spaces isomorphisms are invertible linear maps, in the category of groups isomorphisms are group isomorphisms. This can be confusing. For example \mathbb{R} can be seen as lot of things. It is a set. It is a one dimensional vector space over \mathbb{R}. It is a group under addition. it is a ring. It is a differentiable manifold. It is a Riemannian manifold. In all these \mathbb{R} can be isomorphic (bijective, linearly isomorphic, group isomprhic, ring isomorphic, diffeomorphic, isometric) to different things. This all depends on the context.

**Attribution***Source : Link , Question Author : barf , Answer Author : Thomas Rot*