Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Let $f: X\longrightarrow Y$ and $g: Y\longrightarrow Z.$

Show that $f : X\longrightarrow Y$ is bijective if and only if there exists $g: Y \longrightarrow X$ with $g \circ f = \operatorname{id}_X$ and $f \circ g = \operatorname{id}_Y.$ What happens if you drop one of the two conditions on $g$?

I needed help in constructing simple counterexamples to bijectivity of $f$ in either case.

What I have so far:

Define $g : Y \longrightarrow X$ by $g(f(x)) = x$ for all $x \in X.$ To show that $g$ is well-defined, suppose $f(c)$ and $f(d)$ are the same, but $c$ and $d$ are different (i.e. $f$ isn't injective); that is, let $y = f(c) = f(d),$ where $c \neq d.$ Then $g(y) = g(f(c)) = c,$ and also $g(y) = g(f(d)) = d.$ Hence, $c = d,$ which is a contradiction by the injectivity of $f.$

That $g \circ f = \operatorname{id}_X$ is true by the way we defined $g.$

As for $f \circ g = \operatorname{id}_Y:$

$$\begin{align*} f(g(y)) &= f(g(f(x)) &&\text{for some }x \in X\text{ (since }f\text{ is surjective)}\\ &= f(x) &&\text{(since }g\circ f=\mathrm{id}_X\text{)}\\ &= y \end{align*}$$ as required.

Conversely, suppose there exists $g : Y \longrightarrow X$ with $g \circ f = \operatorname{id}_X$ and $f \circ g = \operatorname{id}_Y.$ Since $f \circ g = \operatorname{id}_Y$ is onto, part a) allows us to conclude that $f$ is surjective (onto). Since $g \circ f = \operatorname{id}_X$ is 1-1, part a) allows us to conclude that $f$ is injective (1-1).

Hence, $f$ is a bijection.

share|cite|improve this question
$\LaTeX$ please. – user38268 Jan 22 '12 at 1:56
Can you please edit it, I am still trying to learn Latex – Buddy Holly Jan 22 '12 at 2:13
Also, you have already posted this question before and it was closed; please see…. Please do not duplicate questions. – Amitesh Datta Jan 22 '12 at 2:27
the only part different is: What happens if you drop one of the two conditions on g ? And this part I'm stuck on – Buddy Holly Jan 22 '12 at 2:29
Try working on this example: $f:\mathbb{Z}\longrightarrow\mathbb{Z},\;f(z)=2z.$ Take $g:\mathbb Z\longrightarrow Z,$ such that $g(2z)=z$ and $g(2z+1)=0$ for any $z\in \mathbb Z.$ Which of the conditions works here? Which doesn't? – user23211 Jan 22 '12 at 2:44

Your argument for why $g$ is well defined is not correct. In fact, it's circular: in order to claim that you can say $g(f(c))=c$, you are already assuming it is well defined, so the contradiction you are arriving at comes from assuming both that $g$ is and is not well defined. You haven't proven that $g$ is well-defined. Moreover, you are assuming that $f$ is not injective, and then you are assuming that $f$ is injective. That's no good.

Instead, let $y\in Y$; since $f$ is onto, there is at least one $c\in X$ such that $f(c)=y$. Since $f$ is one-to-one, there is at most one $c$ such that $f(c)=y$. So for each $y\in Y$ there is one and only one $c\in X$ such that $f(c)=y$. Define $g(y)=c$. You've already worked the "well-definedness" into your definition. The proof that this definition works is fine.

Suppose you have $g\circ f = \mathrm{id}_X$ but not $f\circ g=\mathrm{id}_Y$. By previous parts you know that $f$ is one-to-one. In fact, that's both necessary and sufficient. (HINT: Show that if $g\circ f=\mathrm{id}_X$ and $f\circ h=\mathrm{id}_Y$ for some $g,h\colon Y\to X$, then $g=h$; so the fact that $g$ works as an inverse on one side but not the other will tell you that $f$ cannot be onto).

Something similar if you know $f\circ g = \mathrm{id}_Y$ and we also have $g\circ f\neq \mathrm{id}_X$: now you can conclude one thing about $f$; show that this is the only thing you can conclude, and that the fact that $g$ works as an inverse on the right but not the left tells you the function is not bijective.

share|cite|improve this answer

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.