I am working on the following question.
Let $X$ be a nonempty set and consider a map $f:X\to Y$. Prove that the following are equivalent:
(a) $f$ is injective;
(b) there exists $g:Y\to X$ such that $g∘f=1_{X}$ where $1_{X}:X\to X$ is the identity map;
(c) for any set $Z$ and any maps $h_{1},h_{2}:Z\to X$, the equation $f∘h_{1}=f∘h_{2}$ implies that $h_{1}=h_{2}$.
(*I would like to add here that the "$1$" from each "$1_{X}$" is supposed to be written with the Math Blackboard font - I'm not sure how to do that*)
My issue at the moment arises from being unable to understand (some of) the notation and ideas. I will attempt to describe in my own terms what I believe to understand. I apologize for the terribly imprecise language. You will see why I am not ready to prove anything yet.
(a) That the function $f$ is injective means that it never maps distinct elements of its domain to the same element of its codomain. So every element from the set $X$ maps to a unique element in the set $Y$. And vice versa. A function of the form $f(x)=ax+c$ would be injective (and also a bijection) while a function of the form $f(x)=ax^{2}+bx+c$ would not be (assuming no restrictions on the domain).
(b) This part states that there exists a function $g$ that maps the set $Y$ to the set $X$. The rest I am not sure of. The composite function $g∘f$ equals the identity map $1_{X}$. Does the identity map simply map X back onto X? For instance if $X=\left \{ 1,2 \right \}$, does $1_{X}$ map $1 \to 1$ and $2 \to 2$? So this part would mean, informally, that if you take an element from $Y$, replace it with its corresponding element from $X$, and then map it back to the same element from $X$, you get the identity map. Essentially each element from $Y$ maps to one element from $X$ and you can go back and forth between the sets.
(c) Now the final part. There is a set $Z$ and two maps $h_{1}$, and $h_{2}$, both of which map the set $Z$ to the set $X$. Furthermore (and I dont know if this notation is acceptable),
$f:(h_{1}:Z \to X) \to Y=f:(h_{2}:Z \to X) \to Y.$
I'm not seeing why this shows that $f$ is injective.
I apologize again for this clutter. This is my first real exposure to set theory for my first year Calculus course and our textbook (Stewart) does not cover this. Writing this out has actually helped me to get a better grasp of what I am trying to prove, though I doubt it looks that way. Any help would be appreciated.