# Can a pointwise product of $\mathbb{R}\to\mathbb{R}$ bijections be also a bijection?

It's easy to see that if $f,g:\mathbb{R}\to\mathbb{R}$ are continuous bijections, then their pointwise product $fg$, defined by $x\mapsto f(x)g(x)$ cannot be a bijection.

My question : does it exist a pair $(f,g)$ of bijections from $\mathbb{R}$ to $\mathbb{R}$ such that $fg$ is also a bijection ?

• Why is it impossible for continuous bijections? Commented Feb 6, 2017 at 19:25
• @Alphonse: Since $f,g:\mathbb{R}\to\mathbb{R}$ are continuous bijections, they are strictly monotonic. Hence each of them has infinite limits (of opposite signs) at $-\infty$ and $+\infty$. Multiplying $f$ and $g$ leads to a function whose limits at $-\infty$ and $+\infty$ are both infinite and equal ! This prevents $fg$ from being surjective. Commented Feb 6, 2017 at 19:50

Yes. Our strategy will be to set $g(x)=x$. Furthermore we set $f(0)=0$, and look for some way to make $f$ and $fg$ both bijections on $\mathbb R^\times$.

It will be slightly easier to think instead of the additive group $\mathbb Z_2\oplus \mathbb R$, which is canonically isomorphic to the multiplicative $\mathbb R^\times$. If we let the variable $s$ range over $\mathbb Z_2$, the isomorphism is $(s,x)\mapsto (-1)^s \cdot e^x$.

Now we're looking to find an $f: \mathbb Z_2\oplus \mathbb R \to \mathbb Z_2\oplus \mathbb R$ such that both it and its associate $$\hat f(s,x)\mapsto f(s,x)+(s,x)$$ (which represents $fg$) are bijections.

We will construct $f$ of the form $$f(s,x) = (s,x) + \lambda(s,\lfloor x\rfloor) \qquad (\text{which implies }\hat f(s,x) = (0,2x)+\lambda(s,\lfloor x\rfloor))$$ where $\lambda$ is some function $\mathbb Z_2\oplus\mathbb Z \to \mathbb Z_2\oplus 2\mathbb Z$. (I will let the variable $z$ range over $\mathbb Z_2\oplus\mathbb Z$). This corresponds to saying that $f$ must be just a translation on each half-open interval $[n,n+1)$ for each sign. The requirement that the second component of $\lambda(z)$ must be even guarantees that $\hat f$ will map a half-open unit interval bijectively to $[2m,2m+2)$ for some $m$ with some sign, so we don't risk ending up with an odd-length gap in the image of $\hat f$ to fill.

Now choose an enumeration of $\mathbb Z_2\oplus \mathbb Z=\{z_0,z_1,z_2,z_3,\ldots\}$, and construct $\lambda$ in countably many stages, starting with not having chosen any function values. In stage $k$ we do the following:

1. If $\lambda(z_k)$ has not been defined yet, define it to be the first $z$ (according to the enumeration) that won't make $f$ or $\hat f$ fail to be injective.
2. If $z_k$ is not hit by $f$ yet, find the first $z$ such that $\lambda(z)$ is not yet defined, but can be defined such that $f(z)=z_k$ and without making $\hat f$ non-injective. And define $\lambda(z)$ thus.
3. If $z_k$ has an even second component and is not yet hit by $\hat f$, find the first $z$ such that $\lambda(z)$ is not yet defined, but can be defined such that $\hat f(z)=z_k$ and without making $f$ non-injective. And define $\lambda(z)$ thus.

In each of these steps, there are yet only finitely many values for $\lambda$ chosen, so whenever we choose a $z$ there are at most finitely many that will be excluded -- so even under the additional restriction that either only odd or only even $z$s will work, there will always be one to choose when we need it.

On the other hand, after completing all $\omega$ steps of the construction of $\lambda$, it will be defined for every $z_k$ (due step 1), $f$ will hit every $z_k$ (due to step 2) and $\hat f$ will hit every $z_k$ that is even (due to step 3).

What this adds up to is that $f$ and $\hat f$ are both well-defined and surjective, and they have also explicitly been picked to be injective. So they are bijections!

OLD ANSWER, ASSUMING THE AXIOM OF CHOICE

Lemma. Let $(G,*)$ be any countably infinite group, and let $A$ be a finite subset of $G$. Then there is a bijection $\sigma: G\setminus A\to G\setminus A$ such that $\tilde\sigma(x) = x*\sigma(x)$ is also a bijection $G\setminus A\to G\setminus A$.

Proof by a variant of the "back-and-forth" construction: Let $G\setminus A=\{g_0,g_1,g_2,\ldots\}$, and construct $\sigma$ in countably many stages starting with the empty function:

• In stage $3n$, if $\sigma(g_n)$ is not defined yet, choose a value for $\sigma(g_n)$ such that neither $\sigma(g_n)$ nor $\tilde\sigma(g_n)$ has been hit yet. Otherwise leave $\sigma$ unchanged.
• In stage $3n+1$, if $g_n$ is not yet hit by $\sigma$, choose a $g$ such that $\sigma(g)$ is not yet defined, and $g*g_n$ is not yet hit by $\tilde\sigma$, and define $\sigma(g)=g_n$. Otherwise leave $\sigma$ unchanged.
• In stage $3n+2$, if $g_n$ is not yet hit by $\tilde\sigma$, choose a $g$ such that $\sigma(g)$ is not yet defined and $g^{-1}*g_n$ is not yet hit by $\sigma$, and define $\sigma(g)=g^{-1}*g_n$. (Then $\tilde\sigma(g)=g_n$). Otherwise leave $\sigma$ unchanged.

In each stage only finitely many values of $\sigma$ have been choosen yet, so there are only finitely many choices that cannot be used After $\omega$ many steps $\sigma$ will be as promised in the lemma.

Using the lemma. Define $a\sim b$ if $|a|$ and $|b|$ are rational powers of each other. This is an equivalence relation on $\mathbb R\setminus\{-1,0,1\}$, and each equivalence class together with $\{-1,1\}$ is a countable subgroup of $\mathbb R^\times$.

Define $f(x)$ as follows: First pick one of the subgroups and use the lemma on that subgroup with $A=\varnothing$. Then for each other subgroup, use the lemma on that subgroup with $A=\{-1,1\}$.

Now each nonzero real is defined by exactly one of the $\sigma$s. Let $f$ be the union of these maps, and define $f(0)=0$. This naturally produces a bijection.

Now, with $g(x)=x$ we have that $x\mapsto f(x)g(x)$ is a bijection because it is the union of the $\tilde\sigma$s and $\{0\mapsto 0\}$.

(The lemma does not depend on choice: it can be made deterministic once an enumeration of $G$ has been chosen. But the Axiom of Choice is needed to choose enumerations of all the uncountably many subgroups at once).

(Alternatively, more or less the same proof of the lemma works without the assumption that $G$ is countable, if we start by indexing $G$ by an initial ordinal. Then we don't need to split $\mathbb R^\times$ into subgroups. Of course, the fact that $G=\mathbb R^\times$ can be indexed by an initial ordinal also assumes Choice).

• Thank u Henning ! I am still wondering if there is a way to find a constructive solution to this problem, avoiding anything related to AC. Commented Jan 11, 2017 at 20:33
• @Adren: It turned out that this solution can indeed be modified to be constructive. We can even get an $f$ with countably many discontinuities. Edited. Commented Jan 11, 2017 at 22:32
• With the Axiom of Choice, every bijection from an infinite field to itself can be expressed as the product of two bijections.
– bof
Commented Jan 12, 2017 at 2:31
• @bof: Yes; if only one bijection from the field to itself can be expressed as the product of two bijections, we can get every other bijection in that form too, by precomposing those two factors by an appropriate correction permutation. Commented Jan 12, 2017 at 10:05
• And it seems to be relatively easy to adapt the construction to see that every injection from an infinite field into itself which hits $0$ can be expressed as the product of two bijections. Commented Jan 12, 2017 at 10:14