$g \circ f$ is surjective but $f$ is not surjective. With $f,g$ from $\mathbb{R} \to \mathbb{R} $

There is a similar question If $g \circ f$ is surjective, show that $f$ does not have to be surjective? but it does not my answer my question since it does not require having $f,g$ from $\mathbb{R} \to \mathbb{R} $

I tried many functions but can't find any functions to satisfy the conditions. I am not sure which two functions can be used in this which also have the domain $\mathbb{R}$, for example $log x$ would work in for subjectivity but it does not match the domain.

  • 1
    $\begingroup$ The example in the question you linked is wrong. It has $g(f(x))=|x|$ but the idea is in the right direction. $\endgroup$ – Ross Millikan Nov 14 '17 at 5:50

Define $g(x)=\tan x$ for $x\ne k\pi+\pi/2$, and $g(k\pi+\pi/2)=0$, $k=1,2,...$, and $f(x)=\tan^{-1}(x)$, so $f$ maps $\bf{R}$ onto $(-\pi/2,\pi/2)$, $f$ is not surjective but $g\circ f$ is.

  • $\begingroup$ Another example: $f(x)= \exp \{x\}$, $g(x)=\log \{x\}$ for $x>0$, arbitrary for other x. This is simpler but g is not continuous. $\endgroup$ – Kabo Murphy Nov 14 '17 at 5:50

Let $f$ be a bijection that takes all of $\Bbb R$ to the positive reals, then let $g$ be its inverse. Then $g \circ f$ is the identity, so is surjective.

Alternately, let $$ f(x)=\begin {cases} x &x\not \in \Bbb N\\x+1 & x \in \Bbb N \end {cases}$$ If you think $0 \in \Bbb N$ this covers all of $\Bbb R$ except $0$ so is not surjective onto $\Bbb R$. Again. let $g$ be the inverse. $$g(x)=\begin {cases} x& x+1 \not \in \Bbb N\\x-1 & x+1 \in \Bbb N \end {cases}$$ Again $g \circ f$ is the identity and is surjective on $\Bbb R$. It is a bit of a cheat. $f$ is surjective onto $\Bbb R\setminus \{0\}$. It depends on what set you take for the range of $f$ whether it is surjective. Another (silly) example is $f(x)=x, g(x)=x$ but $f$ is from $\Bbb R$ to $\Bbb R \cup \{cat\}$ so it is not surjective.


A discrete example:

$f: \mathbb{Z} \rightarrow \mathbb{Z}.$

$f(k) = 0$ for $k=0,-1,-2,-3, ...$(neg. integers).

$f(k) = k$ for $k=1,2,3,..$ $ $(pos. integers).

$g: \mathbb{Z} \rightarrow \mathbb{Z}.$

$g(l) = l,$ for $l=-1,-2,-3,.......$(neg. integers.)

$g(m)$ for $m =0,1,2,3.....$is specified below.

$g\circ f(k) = 0$ for $k=0;$

$g\circ f(k):$

$1\mapsto 1$, $2 \mapsto -1$, $3 \mapsto 2,$ $4 \mapsto -2, .....$, etc.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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