Suppose we have $f:A\rightarrow B$ and $g:B\rightarrow C$
if $g\circ f$ is surjective then $g$ is surjective.
This seems one directional as in $\Rightarrow$, but my Professor said if it is a definition then it is always an $\iff$ no matter how the statement is said (or that is how I am interpreting his words, correct me if I am wrong!)
Proof:
Let $c\in C$ be arbritrary. Given that $g\circ f$ is surjective, there exists an $a\in A$ such that $g\circ f(a)= g(f(a))=c$. This implies we have some $b=f(a)\in B$ such that $g(b)= c$. Hence we have shown $g$ is surjective.
But I am wondering is it necessarily true if we try to prove the backwards direction? I seem to run into the problem that I am uncertain if $B$ in $f$ is the same $B$ in $g$.
Best Answer
Your proof is correct. The converse is not true. For example, $g: \{1,2\} \rightarrow \{1,2\}, 1 \mapsto 1, 2\mapsto 2$ is surjective, but if $f: \{1\} \rightarrow \{1\}$ is defined by $1 \mapsto 1$ then $g \circ f$ is not surjective.
Probably what your professor meant was that when we define something, we sometimes use "if" and let "only if" be implied. e.g.,
This is a definition, so really "if" means "if and only if". The reason is that a definition must be an "if and only if" condition to be sensible; otherwise it is ill-defined. Context clues tell you it is a definition: the use of "we say that" and italics around the term surjective (and perhaps that the term has not been used before in the text.) In your case, the statement is not a definition, so the converse is not implied.