Let $f : X \to Y$ and $g: Y \to Z$ be two functions, such that $(g \circ f) : X \to Z$ is bijective

Prove or refute that $f$ is surjective.


For $f$ to be surjective, $\forall b \in Y, \exists a \in X : f(a) = b$

Because $(g \circ f)$ is bijective, the sets X and Z have to have the same number of elements, otherwise the function composition wouldn’t be bijective.

The set Y can have the same number of elements than X and Z or more. It can’t have less, because given that $g \circ f $ is bijective, in other words that X and Z have the same number of elements, the elements in X can map to a smaller number of elements in Y, but the elements in Y can’t map to more elements in Z than there are in Y.

So the function $f$ isn’t necessarily surjective. It could be, but it isn’s necessarily the case.

Are my thoughts correct or not ? If not, why ?? Also, if I am right, how could I write a mathematical proof ?

Thanks for your help !

(P.S. : English isn't my mother tongue, sorry if I made a grammar mistake)

  • 3
    $\begingroup$ It would be better to come up with a very specific and explicit counterexample. Hint: One such counterexample happens with $X=\{1\}, Y=\{1,2\}$ and something appropriate for $Z$ and an appropriate choice for $f$ and $g$. $\endgroup$ – JMoravitz Jul 6 '17 at 19:06
  • $\begingroup$ Can you make an example of sets $X,Y,Z $ and functions $f ,g $ where$f $ is not surjective? $\endgroup$ – Mark S. Jul 6 '17 at 19:07
  • $\begingroup$ If X={1}, Z = {1} but Y = {1,2}, and $f$ and $g$ both map to the same number as the input (so it maps from 1 to 1, 2 to 2, and so on), then $f$ isn't surjective. Is that correct ? $\endgroup$ – Poujh Jul 6 '17 at 19:19
  • $\begingroup$ Yes, that's right. $\endgroup$ – Mundron Schmidt Jul 6 '17 at 19:25
  • $\begingroup$ @Poujh it is almost correct, but the way you have written it (in particular saying it maps 2 to 2) is misleading since $2$ is not in the codomain of $g$. The only possible image in $g$ is the only element in $Z$, namely $1$. The example I was trying to lead you towards is where $f(x)=1$ for all $x$ and $g(x)=1$ for all $x$. $\endgroup$ – JMoravitz Jul 6 '17 at 22:15

Your thoughts are not bad at all, but it seems that you think of $X$, $Y$ and $Z$ as finite sets. This is enough to give a counterexample. You realized that $Y$ has to have at least the same "number" of elements as $X$. But what if $Y$ has more elements? Can $f$ be surjective?

In general, you have to be careful with

Because $(g\circ f)$ is bijective, the sets $X$ and $Z$ have to have the same number of elements, otherwise the function composition wouldn’t be bijective.

Define $A=[0,\infty)$ and $B=[1,\infty)$ and $$ h:A\to B,~h(a)=a+1. $$ Then $h$ is bijective although $B\subsetneq A$.


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