Prove that if $A \preceq B$ and $B \approx C$ then $A \preceq C$

Question

Could anyone check my working please?

Prove that if $A \preceq B$ and $B \approx C$ then $A \preceq C$

Assume $A \preceq B$ (ie. there is a $f:A\to B$ such that it is one-one) and $B \approx C$ (ie. there is a $g:B \to C$ that is one-one and onto), we need to show that there is a $h:A\to C$ that is one-one.

Let this $h$ be $g\circ f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.

We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.

Answer 1

If you want to show $g \circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.

Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.