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!

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    $\begingroup$ The theorem also says they both have to be $F$-vector spaces for $F$ some fixed field. I mean, $\mathbb{F}_2^2\not\cong \mathbb{R}^2$... $\endgroup$
    – user1729
    Jul 28, 2011 at 10:23
  • $\begingroup$ This somewhat reminds me of this: math.stackexchange.com/questions/27871/… $\endgroup$
    – Asaf Karagila
    Jul 28, 2011 at 11:03
  • $\begingroup$ I'm not following a lot of this...forgetting categories and algebraic structures, isn't an isomorphism between two (unstructured) sets a bijection? And if you are trying to preserve functional relationships, then you're talking about a homomorphism? Isn't -that- the level of terminology the OP is asking about? $\endgroup$
    – Mitch
    Jul 29, 2011 at 2:08

5 Answers 5


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.

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    $\begingroup$ It may seem picky, but I think you are saying things somewhat misleadingly in your second paragraph. The definition of an isomorphism in a concrete category is not a bijective morphism. (The definition of isomorphism in any category is a morphism with a two-sided inverse.) In many "algebraic" concrete categories it turns to be true that a morphism is an isomorphism iff it is bijective, but this is not true in the category of topological spaces: a continuous bijection need not be an isomorphism. $\endgroup$ Jul 28, 2011 at 10:51
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    $\begingroup$ In fact your definition in the category of differentiable manifolds is also incorrect: $x \mapsto x^3$ is a differentiable bijection from $\mathbb{R}$ to $\mathbb{R}$ but not a diffeomorphism. $\endgroup$ Jul 28, 2011 at 10:54
  • $\begingroup$ @Pete thanks, I was way too sloppy there. $\endgroup$
    – Thomas Rot
    Jul 28, 2011 at 11:31
  • $\begingroup$ Thanks to Thomas and Tobias. This helped a great deal. $\endgroup$
    – barf
    Jul 29, 2011 at 7:11

The difference is that an isomorphism is not just any bijective map. It must be a bijective linear map (ie, it must preserve the addition and scalar multiplication of the vector space).

So even though you have a bijection between $\mathbb{R}$ and $\mathbb{R}^2$, there is no way to make this bijection linear (as follows from looking at their dimensions)

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    $\begingroup$ May I emphasize that the idea of a bijection can be applied to any type of set A and B whereas isomorphisms need that A and B are vector spaces. $\endgroup$
    – Listing
    Jul 28, 2011 at 10:01
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    $\begingroup$ Well, the term isomorphism depends on the category in which one works (as mentioned in Thomas's answer). This means that it can be used in many contexts. $\endgroup$ Jul 28, 2011 at 10:10
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    $\begingroup$ Ok fine, I just wanted to mention that an isomorphism is closely related to the structure of the sets you are working with while a bijection is a very general property of a map. $\endgroup$
    – Listing
    Jul 28, 2011 at 10:24
  • $\begingroup$ @TobiasKildetoft: Thanks for help. $\endgroup$
    – Mikasa
    Mar 11, 2013 at 13:38

First of all, let me give a general definition of isomorphism. In a context where you have been given a notion of morphism between two objects, as well as a notion of an identity morphism, an isomorphism is simply a morphism $f : A \to B$ such that there exists a morphism $g : B \to A$ where we have composites $f \circ g = \mathrm{id}_B$ and $g \circ f = \mathrm{id}_A$; such a $g$ is called an inverse morphism.

What does this mean in the context of sets? Well, a morphism here is just any function, and the identity morphism on a set $A$ is the function which maps $x \mapsto x$. So it can be shown that an isomorphism in this category of objects is just a bijection of sets.

What about in linear algebra? The objects in question are vector spaces over some common field $K$, and the morphisms are the $K$-linear maps. An isomorphism, then, is a linear map which admits an inverse linear map. It turns out that a linear map $f : A \to B$ to has an inverse if and only if $f$ is bijective. But it is not enough for two vector spaces to have the same cardinality, since there may be no bijections which are also linear maps! (This is the crux of your confusion, it seems.)

I will also mention now that there are categories in which bijective morphisms are not isomorphisms. For example, consider the category of topological spaces and continuous maps. Then, there is a bijective continuous map from the half-open unit interval $[0, 1)$ to the circle $S^1$... but there is no bijective continuous map from $S^1$ to $[0, 1)$, so these two are not isomorphic as topological spaces.


An isomorphism (of vector spaces) is a bijective linear transformation between these spaces that preserves addition and scalar multiplication.Now, as you correctly said, $|\mathbb{R}| = |\mathbb{R}^2|$, so there certainly exist bijections between $\mathbb{R}$ and $\mathbb{R}^2$. The question is, do any of these bijections preserve addition and scalar multiplication? Of course, it's impossible to check this directly; I mean, it's impossible to write down all these bijections and check if any one of them preserves addition and scalar multiplication. So instead, we use the theorem you mentioned: two vector spaces are isomorphic if and only if they have the same dimension. So since $\mathbb{R}$ has dimension $1$ (as a vector space over $\mathbb{R}$ and $\mathbb{R}^2$ has dimension $2$ (again, as a vector space over $\mathbb{R}$) so it follows that there is no isomorphism between $\mathbb{R}$ and $\mathbb{R}^2$. In other words, even though there exist bijections between $\mathbb{R}$ and $\mathbb{R}^2$, none of them will preserve addition and/or scalar multiplication. As a final remark, I should mention that if you consider $\mathbb{R}$ and $\mathbb{R}^2$ as vector spaces over $\mathbb{Q}$, then $\mathbb{R}$ and $\mathbb{R}^2$ will be isomorphic since both will have dimension $\mathfrak{c}$. So the statements "$\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as vector spaces" or "$\mathbb{R}$ and $\mathbb{R}^2$ are not isomorphic as vector spaces" are ambiguous; you need to specify the field over which you are working (whether it is $\mathbb{Q}$ or $\mathbb{R}$ or something else).

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    $\begingroup$ In the last sentence, the word "ambiguous" seems preferable to "meaningless": we have too many possible meanings, not too few. $\endgroup$ Jul 28, 2011 at 10:59
  • $\begingroup$ @Pete L.Clark Thanks! I have changed "meaningless" to "ambiguous" $\endgroup$ Jul 28, 2011 at 18:04

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).


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