Exercise 3.3.4 (Cancellation laws for composition)

For the first claim, suppose $f\ne \tilde{f}$. Then there exists some $x\in X$ such that $f(x)\ne \tilde{f}(x)$. Since $g$ is injective, this implies that $(g\circ f)(x)\ne (g\circ \tilde{f})(x)$ which contradicts the assumption that $g\circ f=g\circ \tilde{f}$. If $g$ is not injective, this statement is not true. As a counter example, let $c\in Z$ and consider $g$ such that $g(y)=c$ for all $y\in Y$ (assuming $Y\ne \{y\}$). This $g$ is not injective and clearly $g\circ f=g\circ \tilde{f}$ for any $f$ and $\tilde{f}$ considered, including where $f\ne \tilde{f}$.

For the second claim, suppose $g\ne \tilde{g}$. This implies that there exists a $y\in Y$ such that $g(y)\ne \tilde{g}(y)$. Since $f$ is surjective, there exists $x\in X$ such that $f(x)=y$. So we have $(g\circ f)(x)\ne (\tilde{g}\circ f)(x)$ which contradicts the assumption that $g\circ f=\tilde{g}\circ f$. If $f$ is not surjective this statement is not true. As a counter example, consider $f$ a constant function, i.e. let $f(x)=a$, $a\in Y$ (again assuming $Y\ne \{a\}$) for all $x\in X$. Then, $f$ is not surjective. Then, $g\circ f=\tilde{g}\circ f$ implies $g$ and $\tilde{g}$ are equal at $a$, i.e. $g(a)=\tilde{g}(a)$, but can differ at any other point $a^{\prime }\in Y$, $a\ne a^{\prime }$ and so does not imply that $g=\tilde{g}$.