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Let $A$ and $B$ be (non-empty) sets and $f:A\rightarrow B$. Prove that the following statements are equivalent.

a) $f$ is one-to-one.

b) There is a function $g:B\rightarrow A$ that satisfies $(g \circ f)=I_A$.

c) If $h:C\rightarrow A$ and $k:C\rightarrow A$ satisfy $(f \circ h)=(f \circ k)$ then $h=k$.

Apologies in advance if my formatting is not ideal. Note "o"=composition

Thanks!!

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    What parts have you done so far? – 2017-01-20
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    @DonAntonio I'm struggling with it overall. I know my definitions, just confused how to connect them. – 2017-01-20

2 Answers 2

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a) $\implies$ b)

Since $f: A \to B$ is an one-to-one, it is a bijection onto its image. Hence, there is an inverse $h: f(A) \to A$. Choose a point $x \in A$. Define $g: B \to A$ by

$$ g(y) = \begin{cases} h(y) && y \in f(A) \\ x && \mbox{else} \end{cases} $$

Then $g\circ f = I_A$.

Answer from Left Inverse: An Analysis on Injectivity

b) $\implies$ c)

Suppose we have $f\circ h = f\circ k$. Then, applying $g$ from b) to both sides, we get $$g \circ f\circ h = g\circ f\circ k$$ $$h = k$$

c) $\implies$ a)

Assume $f(a) = f(b)$ with $a \neq b$. Then define $k,h: A \rightarrow A$ with $$k(x) := \begin{cases} b &&x=a \\ a &&x = b\\ x &&\mbox{ else}\end{cases}.$$ Take for $h:= I_A$. Then we have $f\circ k =f \circ h$ but $f \neq h$.

Answer from: How to prove that monos are injective?

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    Did you use the "axiom of choice" method for a=>b? – 2017-01-20
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    I don't believe so, no. But I think you need to use the axiom of choice to prove that if a function is onto (i.e. surjective) then it has a right-inverse – 2017-01-20
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    I'm confused about the c)==>a) part of the proof – 2017-01-21
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    What's $f(g(a))$? From how we defined $g:A\rightarrow A$, it's $f(b)$. As we assumed $f(a) = f(b)$, we get that $f(g(a)) = f(a).$ Similarly, $f(g(b)) = f(b)$ (check this!). As $h$ is the identity on $A$, $f(h(a)) = f(a)$ and $f(h(b)) = f(b)$. For any other point $x$ in $A$, $f(g(x)) = f(x)$ and $f(h(x)) = f(x)$. So $f\circ g = f\circ h$, because the two functions agree on all points in their domain. However, $g$ is not equal to $h$, because $g(a) = b \neq a = h(a)$. Feel free to ask if this isn't clear – 2017-01-22
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    I understand what you did here. But how does this show that f is one-to-one? – 2017-01-22
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    The definition of one-to-one is that whenever $f(a) = f(b)$, we have $a=b$. Here, we started by assuming $f(a) = f(b)$ and $a\neq b$ and then arrived at a contradiction. Hence, our assumption must have been wrong and so we must've had $f(a) = f(b)$ and $a=b$ – 2017-01-22
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    okay that clears things up a bit. I think I'm still mostly confused because in the question statement c) is using function k and h from C-->A, and I don't see this anywhere in the proof – 2017-01-23
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    The statement of c) is simply saying that if $h$ and $k$ have the same domain ("$C$"), then whenever you have $f\circ h = f\circ k$, you must have $h = k$. In our example, the two functions do indeed have the same domain: $A$ (which corresponds to $C$ in the definition). – 2017-01-23
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    THank you, I understand that part now. Just clarifying in part c)-->a), what the domain for x is? I think its domain is A? – 2017-01-24
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    Do you mean the domain of $f$? If so, then yes, that's $A$. – 2017-01-24
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    I mean the x in the definition for function k(x). Where k(x)=x when x=else – 2017-01-24
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    Oh, ok. Well yeah, when we write $k(x)$ we're gonna want $x$ to be in $k$'s domain, which is $A$ – 2017-01-24
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    Makes sense. Thank you for all the help – 2017-01-24
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$1)\Rightarrow 2)$ $\forall y\in B$ then $y\in Im(f)$ or $y\notin Im(f)$, if $y\in Im(f)$ then $\exists x_y \in A; f(x_y)=y$

let define $g:B\to A$ by form $g(y)=x; y \notin I(f)$($x$ fixed point ) or $g(y)=x_y; y \in I(f)$ by definition $g$ we have $domain(g)=B$. Let be $y,y'\in B; y=y'$ then

case$(1)$: if $y,y'\in I(f)$ then $\exists x_y,x_{y'}\in A; f(x_y)=y, f(x_{y'})=y'$ then $g(y)=x_{y},g(y')=x_{y'} $ . we have $f$ is one-to-one and $y=y'$ then $f(x_y)=f(x_{y'})\Rightarrow x_{y}=x_{y'}\Rightarrow g(y)=g(y') $

case$(2)$: if $y,y'\notin I(f)$ then $g(y)=g(y')=x$

in two case we have $g(y)=g(y')$ then $g$ well defined and $g$ is a map and

$(g\circ f)(x_y)=g(f(x_y))=g(y)=x_y=id(x_y)$

$2)\Rightarrow 3)$ we have $h:C\to A$ and $k"C\to A$ by $2)$ there is $t:A\to C; t\circ h=t\circ k=I$, we have $f\circ h=f\circ k\Rightarrow t\circ f\circ h=t\circ f\circ k \Rightarrow h=k$

$3)\Rightarrow 1)$ $f(x)=f(y)\Rightarrow f(h(a))=f(h(b))\Rightarrow fh(a)=fh(b) $ by $3)$ $h(a)=h(b)\Rightarrow x=y$ then $f$ one to one map.