Proving that 3 set statements are equivalent 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!!
 A: 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?
A: $1)\Rightarrow 2)$ $\forall y\in B$ 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. 
