MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello Everybody.

I need some more examples for the following really interesting phenomenon:

   A function from the class ... is one-one iff it is onto. 

Some examples I know:

1) Finite set case: functions from $\lbrace 1,2,\dots,n\rbrace$ to itself is one-one iff onto.

2) Linear operators $T\colon V\rightarrow V,$ where $V$ is a finite-dimensional vector space is also one-one iff onto.

3) Linear operators of the from (I-K) where K is some compact operator acting on a Banach space satisfies this property. This is the famous result of Fredholm.

It is very easy to find domains where the result fails.

I remember my teacher telling me that 'compactness is the next best thing to finiteness', hence this result which trivially holds in the finite case can happen only in the compact setting. I would like to know, if this is really the case or are there any other examples?

Thank you in advance.

EDIT: Looking at some answers, I thought it is better if the scope of the question is broadened.

Does injection (surjection) imply surjection (injection) and isomorphism/isometry? (i.e. by assuming one-one can I get ontoness and structure preserving properties free)

share|cite|improve this question
I heard similar things about compactness being the next best thing after finiteness. I just never understood what is so great about finiteness... – Asaf Karagila Mar 30 '12 at 19:20

16 Answers 16

Tom Leinster was talking about this here

share|cite|improve this answer
Thanks. Very closely related link. The conclusion of the author that we need three different proofs for the same phenomenon is also interesting to ponder over. – Uday Feb 17 '12 at 20:56
Thanks, Moshe. There was a follow-up post (…) unifying some but not all aspects of the three situations. In particular, there's a unified proof that if (injective implies iso) then (surjective implies iso). – Tom Leinster Feb 17 '12 at 22:07
I've expanded on that last fact in my answer on this page. – Tom Leinster Feb 17 '12 at 22:28
In the case of a compact space, it seems that the eventual image is called $\omega$-limit set. – Denis Serre Feb 18 '12 at 9:08

If $A$ is a noetherian ring, then a ring homomorphism $f: A \to A$ is surjective iff it is an isomorphism.

(see the accepted answer of this question)

share|cite|improve this answer
The proof for the (I-K) where K is compact operator(stated in the question) also has a similar proof! The nesting is on kernels. – Uday Feb 17 '12 at 20:24

Uday asked about the double implication "one-to-one iff onto". Here's an observation about the relationship between the two different directions of implication, taken from here.

Let $\mathbf{M}$ be a symmetric monoidal closed category. (This is overkill, but I won't attempt to be more precise.) Suppose we have in $\mathbf{M}$ two distinguished classes of map, called the "injections" and the "surjections", with the following two properties: (i) if $s\colon X \to Y$ is a surjection then for all $Z$, the induced map $$ s^*\colon \mathbf{Hom}(Y, Z) \to \mathbf{Hom}(X, Z) $$ is an injection, and (ii) any surjection with a left inverse is an isomorphism. Suppose, finally, that every injective endomorphism in $\mathbf{M}$ is an isomorphism. Then every surjective endomorphism in $\mathbf{M}$ is an isomorphism.

Proof: let $s\colon X \to X$ be a surjective endomorphism. Then $s^*\colon \mathbf{Hom}(X, X) \to \mathbf{Hom}(X, X)$ is an injective endomorphism, and therefore an isomorphism. It follows that there exists $t\colon X \to X$ such that $s^*(t) = 1_X$, that is, $t\circ s = 1_X$. But then $s$ is a surjection with a left inverse, so $s$ is an isomorphism.

The hypotheses hold in the category of finite sets, the category of finite-dimensional vector spaces, and the category of compact metric spaces. In the last case, the maps are the distance-decreasing maps (in the weak sense), "injection" should be interpreted as "isometry into", and "surjection" has its usual meaning.

In particular, if you already know that every isometry from a compact metric space into itself is surjective, this lets you deduce that every distance-decreasing surjection from a compact metric space to itself is an isometry.

share|cite|improve this answer
@Tom Leinster I liked the way duality of injection and surjection is exploited to bring about a unification. A very good observation indeed! Thank you. (To be frank, I have not understood the observation completely.) – Uday Feb 17 '12 at 22:51
"Symmetric monoidal closed category" just means that you have a collection of objects (e.g. finite-dimensional vector spaces), some maps between them (e.g. linear maps), a way of taking the "product" of two objects (e.g. the tensor product), and a way of taking the "space of maps" between two objects (e.g. the vector space Hom(V, W) of linear maps from V to W). There are some axioms and details to fill in, but that's the idea. – Tom Leinster Feb 17 '12 at 23:42

Although I tend to shy away from list questions, this is more fun to contemplate than the pile of exams on my desk right now.

So take a compact Riemann surface $X$ with genus $g\ge 2$. A nonconstant holomorphic self map $f:X\to X$ is necessarily an isomorphism. Proof: Surjectivity is automatic by, for example, the open mapping theorem. If $f$ were not injective, then it would have a degree $d>1$. But the Riemann-Hurwitz formula would give $g-1\ge d(g-1)$ which is impossible.

share|cite|improve this answer
This cannot be. I think that in your application of Riemann-Hurwitz you are assuming that your map $f$ is everywhere unramified, which is a quite strong assumption! – Tommaso Centeleghe Feb 17 '12 at 22:14
RH says $2g-2 = d(2g-2)+\sum_p (e_p-1)$ which gives the inequality I stated. Tell me which part do you not understand? – Donu Arapura Feb 17 '12 at 22:29
oooops.. you are right. I misread what you wrote, I thought you were considering $f:X\rightarrow Y$..good! – Tommaso Centeleghe Feb 17 '12 at 22:49
OK, no harm done. Let me edit to make clearer. – Donu Arapura Feb 17 '12 at 22:51
Is there no way we can come out of compactness assumption? – Uday Feb 17 '12 at 23:01

(edited slightly) A one to one continuous map between two compact $n$-dimensional manifolds without boundary having equal numbers of components must be onto, and in fact a homeomorphism.

The image of any connected component must be connected, open (by invariance of domain), and closed (by compactness), and therefore must be a component.

share|cite|improve this answer

A surjective endomorphism of a finitely generated residually finite group is an isomorphism.

share|cite|improve this answer

Let $G$ be a sofic group (see this survey article of Pestov for the definition and various results) - all amenable groups are sofic, as are all free groups, and no groups are known to be non-sofic. Let $X=\{0,1\}^G$ with the product topology and let $f: X \to X$ be a continuous function which is also a right $G$-map, here $G$ acts on $X$ by shifts. Then if $f$ is injective, it is automatically surjective - this is Gromov's partial solution to the Gottschalk surjunctivity conjecture, which I think is also mentioned in Pestov's article.

Now this is not what your question asked for, but if we now look at $C(X)$ and the induced algebra homomorphism $f^* : C(X) \to C(X)$, then

$f^*$ surjective $\iff$ $f$ is injective (Tietze/Urysohn) $\iff$ $f$ is bijective (above) $\iff$ $f^*$ bijective (Tietze/Urysohn)

share|cite|improve this answer
This is a good one. Isn't Gromov's partial solution an extension of Ax-Grothendieck to some sort of "limit varieties"? – Jon Bannon Feb 17 '12 at 23:45
Jon, I must confess that I don't know how Gromov's proof works, but given that one can define and investigate sofic groups in terms of embeddings into suitable ultrapowers, your suggestion sounds very plausible. – Yemon Choi Feb 18 '12 at 3:31

Let $R$ be a right perfect ring. Then an endomorphism on a finitely presented left $R$-module is injective if and only if it is surjective: reference.

share|cite|improve this answer

If $A$ is a ring and $M$ is a finitely generated $A$-module, then an $A$-module endomorphism $f:M\rightarrow M$ is surjective iff it is an isomorphism.

share|cite|improve this answer
Thanks. This statement is a direct generalization of finite dimensional vector spaces. May be we can extend the results to infinite dimensional modules and compact operators defined on them. – Uday Feb 17 '12 at 20:44
Is this true if the ring doesn't satisfy the invariant basis number property: Or did you mean to consider only (say) commutative rings? (I am interested in your example; do you have a reference?) – Todd Trimble Feb 17 '12 at 20:53
@Todd Trimble I doubt this may be true in general. In vector space setting, we will need to have a condition of dimension of the range and kernel to be equal. In generalizing to module, this condition may demand IBN. – Uday Feb 17 '12 at 21:11
It's obviously false if IBN fails, but it is true over commutative rings. It's a consequence of (an appropriate form of) Nakayama's lemma. I think you can find it in Atiyah & MacDonald. I have no idea about what happens if commutativity fails but IBN still holds. – Harry Altman Feb 17 '12 at 21:16
@Harry: I should have been more clear that my first question was rhetorical. Thanks for the reference! – Todd Trimble Feb 17 '12 at 21:53

Let $G$ be a discrete group, and let $T:\ell^2(G)\to \ell^2(G)$ be a bounded linear operator which commutes with all right translations: that is, if $\xi\in \ell^2(G)$ and $g\in G$ then $T(\xi\cdot g) =T(\xi)\cdot g$. (In other words, $T$ belongs to the group von Neumann algebra.) Then if $T$ is surjective, it is invertible.

This follows from combining a result of Kaplansky with the fact that $C^\ast$-algebras are inverse-closed in containing $C^\ast$-algebras. My own feeling is that the result is tacit folklore but in any case it follows by duality from Theorem 3.2 in this paper.

share|cite|improve this answer
Thanks. Can we prove that this class has a non-compact operator? – Uday Feb 17 '12 at 21:56
A II_1 von Neumann algebra has no non-zero compact operators (no minimal projections). So yes, this is different from the Fredholm examples. – Yemon Choi Feb 17 '12 at 22:04
In fact, by considering polar decomposition this says that co-isometries are always isometries, which is the defining property of finite von Neumann algebras. – Jesse Peterson Feb 18 '12 at 2:53
Jesse: that's a snappy way to put it, and that's how Kaplansky implicitly puts it. I guess I like to follow in Dixmier's wake by thinking of finite von Neumann algebras as those that have a separating family of finite normal traces. This seems to generalize better to situations where VN(G) is not finite but where the unitization of $C_r^\ast(G)$ has the "left-invertible imples invertible" property, see my older MO question… – Yemon Choi Feb 18 '12 at 3:30

If $f : S \to S$ is volume preserving and $S \subseteq \mathbb{R}^n$ has finite volume, then f is injective iff f is surjective.

There is a very elegant proof of Koebe–Andreev–Thurston theorem (given in the book on combinatorial geometry by Agarwal and Pach) using this property.

share|cite|improve this answer
Erik, I couldn't understand your first sentence, at the grammatical level. Could you rephrase? – Tom Leinster Feb 17 '12 at 22:12
@Tom : oops, I've fixed it now. (The intention was to answer the question "under what circumstances is f injective iff f is surjective", but I just realized the question was not given like that.) – Erik Aas Feb 18 '12 at 9:06
Got it! Thanks. – Tom Leinster Feb 18 '12 at 15:23
Actually, I don't get it. Take n = 1 and S = [0, 1]. Take the map f: S --> S defined by f(x) = 2x (mod 1). Then f is measure-preserving (wrt the usual measure) and surjective, but not injective. Maybe you mean something different by "volume-preserving"...? I had a look in Agarwal and Pach's book, but couldn't find a clean statement of the result to which you're alluding. – Tom Leinster Feb 20 '12 at 11:40
Does $f$ have to be continuous, or something like that? Otherwise, the result seems to be trivially false because you can mess about with the map on a set of measure zero. – gowers Mar 31 '12 at 14:04

This addresses the "broader scope" of the question and possibly the comments of Uday on Donu's answer: An injective morphism from an affine algebraic variety over an algebraically closed field to itself is also surjective. Moreover, probably even more surprising is the fact that in the case that the field has characteristic zero (and of course algebraically closed), an injective endomorphism is actually a polynomial automorphism (that is the inverse is also a polynomial map!). See e.g. Chapter 4 of van den Essen's "Polynomial Automorphisms" for proofs of both these statements. Also from the same book: the map $x \mapsto x^3$ from $\mathbb{Q} \to \mathbb{Q}$ shows the necessity of algebraic closedness of the field, and the Frobenius automorphism $x \mapsto x^p$ of an algebraically closed field of characteristic $p > 0$ shows that the second statement is false for positive characteristic. Also, note that both statements are automatically true for proper varieties.

share|cite|improve this answer

The multiplication maps of modules over an Artin Ring have this property. Artin rings similarly generalize both finite sets and vectors spaces.

share|cite|improve this answer

If $G \subset \mathbb{C}^n$ is a domain and $f: G \mapsto \mathbb{C}^n$ is an injective holomorphic mapping, then also $f(G)$ is a domain and $f: G \mapsto f(G)$ is biholomorphic.

This follows from the fact that, under the same assumptions, if $f$ is injective, then det $J_f(z) \ne 0$ for every $z \in G$. ($J_f$ denotes the complex jacobian.)

share|cite|improve this answer

The Dixmier conjecture involves an even stronger statement. Let $A_n$ be the Weyl algebra, which is the algebra of polynomial differential operators on $\mathbb C[x_1,\ldots,x_n]$. The conjecture is that any algebra map $f:A_n \to A_n$ is an isomorphism. (Of course, it isn't hard to see that $A_n$ has no two-sided ideals, so any such $f$ is automatically injective.)

It was recently proved that the Dixmier conjecture is stably equivalent to the Jacobian conjecture, in the sense that if one conjecture is true for all $n$, then so is the other. (References are given in the wikipedia page.)

share|cite|improve this answer

An isometry (i.e. distance preserving map) between metric spaces is automatically injective.

share|cite|improve this answer
How does this address the original question? – Yemon Choi Feb 17 '12 at 22:18
Much more interestingly, a distance-decreasing function between compact metric spaces is a surjection iff it is an isometry. See the discussion in the thread linked to by Moshe. – Todd Trimble Feb 17 '12 at 23:21
@Yemon Choi: it doesn't really relate to the question. sorry for this disattention. and thanks for warning. – Hans Feb 18 '12 at 13:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.