Let $A, B$ be non-empty sets and $f\colon A\to B$ be a function. Let $C$ be a set with at least two elements. Denote by $\text{Map}(A,C)$ the set of all functions from A to C. Denote by $\text{Map}(B,C)$ the set of all functions from $B$ to $C$.
Define $f^{*}\colon \text{Map}(B,C)\to \text{Map}(A,C)$ by $f^{*}(g)= g \circ f$ for any $g\in\text{Map}(B,C)$
Consider the statements below:
(A) The function $f$ is surjective.
(B) The function $f^{*}$ is injective.
a) Suppose A holds. Does B hold then?
b) Suppose B holds. Does A hold then?
Thanks.
Properties (A) and (B) are equivalent.
Suppose (A) holds true and let $f^*(g)=f^*(h)$. Then $g\circ f=h\circ f$. Let $y\in B$ arbitrary. By the surjectivity of $f$ there is some $x\in A$ such that $f(x)=y$. So $g(y)=g(f(x))=h(f(x)=h(y)$, i.e. $g=h$.
Now assume (B) and that $f$ is not surjective. Define $g(x)=c$ for all $x\in B$ with $c\in C$. By assumption there is some $d\in C$ different from $c$. Define $h$ by $h(x)=c$ for $x\in f(A)$ and by $h(x)=d$ If $x\in B\setminus f(A)$. Then $g\not=h$ but $f^*(g)=f^*(h)$, a contradiction.