I want to check with you my proof of the following result:
Result: Let $f:A \rightarrow B$ and $g:B \rightarrow C$ be functions and let $U \subseteq A$. Then $(g \circ f)_*(U) = g_*(f_*(U))$.
My proof: In order to show that $(g \circ f)_*(U) = g_*(f_*(U))$, it suffices to show that each of the sets is a subset of the other.
Let $x \in (g \circ f)_*(U)$. By definition of image, we deduce that exists some $u \in U$ such that $x = (g \circ f)(u)$. By definition of composition, we observe that $(g \circ f)(u) = g(f(u))$, so $x = g(f(u))$. Again, by definition of image, we know that $g(f(u)) \in g_*(f_*(U))$. Hence $x \in g_*(f_*(U))$. Therefore $(g \circ f)(U) \subseteq g_*(f_*(U))$.
Now, let $y \in g_*(f_*(U))$. By definition, there exists some $a \in f_*(U)$ such that $y=g(a)$. Let $a_0$ be that element. Since $a_0 \in f_*(U)$, we deduce that there exists some $b \in U$ such that $f(b)=a_0$. Let $b_0$ be that element. Hence $f(b_0)=a_0$ and $g(a_0)=y$. Then for $b_0 \in U$ we have $g(f(b_0)) = y$. By definition, $g(f(b_0))=(g \circ f)(b_0)$. So $g(f(b_0)) \in (g \circ f)_*(U)$. Hence $y \in (g \circ f)_*(U)$. Therefore, we have that $g_*(f_*(U)) \subseteq (g \circ f)_*(U)$.
This concludes our proof.
Any feedback on this proof is welcome! Thank you!