Prove: $f$ is injective if and only if for any non-empty set $C$ and for all functions $g, h : C \to A$ such that $f ◦ g = f ◦ h$ we have $g = h$.

Question

I’m attempting to prove this exercise for my elementary analysis class, however, I am having difficulty deconstructing the iff statement. Any help in understanding how the implications are working would be great. I’m also lost on how $f$ being injective makes the functions $g$ and $h$ equal. Thanks in advance.

Answer 1

Hints:

So, you want to prove that $f$ being injective implies that, whenever $f\circ g = f \circ h$, you will consequently have $g=h$. You also want to prove the converse: whenever you have $f\circ g = f\circ h \implies g = h$, you want to show that this implies $f$ is an injective function.

For the forward implication, one of the basic properties injective functions have is the existence of a left inverse. That is, if $f : A\to B$ is injective, there exists a function $\bar f : B\to A$ such that $\bar f \circ f = 1_A$. With this in mind, try to see if you can make this fact useful.

For the converse, recall the definition of injective functions. One of the definitions you may have is that $f$ is injective if $f(x) = f(y) \implies x = y$ for all $x,y$ in the domain of $f$. We're given that $f(g(x)) = f(h(x))$ for all $x$ in the domains of $g,h$, and that $g(x)=h(x)$ is implied as a result. How might you use this fact when it comes to the definition of injective?

Answer 2

One way is obvious. If $f$ is injective then the condition holds by definition. to prove the converse suppose $f(x)=f(y)$. We want to show that $x=y$. We have to choose appropriate $C,g$ and $h$ and use the hypothesis. Note that if $f(g(z))=x$ and $f(h(z))=y$ for all $z \in C$ we get $x=y$ as required. For this we take $C=\{x,y\}$ and define $g$ to be the constant function $x$ and $h$ to be the constant function $y$.