# If $h$ is an injection, with $hf=hg$, then $f=g$

Let $A,B,C$ be nonempty sets and let $f,g \in \mathbb{F}(A,B)$, the set of all functions from $A \to B$. Suppose $h \in \mathbb{F}(B,C)$ is injective, and $hf=hg$. Show that $f=g$. (Note that $hf=h(f(x))$ with my textbook's notation.)

I know that by the definition of injective, this means that $f(h_1)=f(h_2) \iff h_1=h_2$, But I don't know how this applies to the proof. Is other some other property of injections that I am ignoring, or do I need to start somewhere else first? A hint on how to get started would be very helpful.

1. Take any $a\in A$.
2. Compare the values $h(f(a))$ and $h(g(a))$ (use the fact that $hf=hg$)
3. Use the fact that $h$ is injective, together with what you figure out from 2.
4. Conclude that $f(a)=g(a)$