Suppose $\displaystyle f$, $\displaystyle g$, and $\displaystyle h$ are all mappings of a set $\displaystyle A$ into itself.

a) Prove that if $\displaystyle g$ is onto and $\displaystyle f \circ g = h \circ g$, then $\displaystyle f = h$

By Contradiction:

If $\displaystyle g$ is onto and $\displaystyle f \circ g=h \circ g$; then $\displaystyle f \neq h$

Proof:

let $\displaystyle f:A \rightarrow A$ and $\displaystyle g:A \rightarrow A$

so the composition of $\displaystyle f \circ g : A \rightarrow A$

Suppose $\displaystyle \exists a \in A$

since $\displaystyle g$ is onto,

$\displaystyle \exists b \in A \mid g(b)=a$

By definition $\displaystyle f \circ g = h \circ g$

thus $\displaystyle f(a)=f(g(b))=h(g(b))$

therefore f = h and we arrive at a contradiction

I corrected a few of the things you pointed out does that look better? The way our professor likes stuff is similar to the way I wrote it out here so for consistency I just try to learn it that way ... helps for regurgitation

as for part b what do you think about the following

for an arbitrary x in a .. by definition f(g(x)) = f(h(x)) and f is one-to-one then g(x) = h(x) so g must equal h

how does that look logic wise?