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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.

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    $\LaTeX$ please.2012-01-22
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    Can you please edit it, I am still trying to learn Latex2012-01-22
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    Also, you have already posted this question before and it was closed; please see http://math.stackexchange.com/questions/100102/one-to-one-and-onto-functions. Please do not duplicate questions.2012-01-22
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    the only part different is: What happens if you drop one of the two conditions on g ? And this part I'm stuck on2012-01-22
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    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?2012-01-22
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    @Amitesh: I don't see how this is a duplicate of the question you linked to, even without the addition that Buddy pointed out. Of course they're closely related, but I think seeing how they're related requires roughly as much insight as solving either of them would.2012-01-22
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    @joriki My comment referred to the first version of the question; the question was edited after my comment and (I think) your comment was posted after this revision. I agree that the current revision of the question is not a duplicate of the question I linked (but the first revision was).2012-01-22
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    For a more interesting example, see http://math.stackexchange.com/questions/70777/a-ring-element-with-a-left-inverse-but-no-right-inverse/70781#707812012-01-24

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