I am taking a class on set theory and am having troubles with proofs. Here is a question I am currently doing:
Let $f : A \rightarrow B$ be 1-1. Prove that $\forall$ set $S \forall g_1, g_2 : S \rightarrow A, f \circ g_1 = f \circ g_2 \Rightarrow g_1 = g_2$.
So far, this is what I have written:
$a_1 = g_1(s_1)$, where $a_1 \in A$ and $s_1 \in S$
$a_2 = g_2(s_2)$, where $a_2 \in A$ and $s_2 \in S$
$f \circ g_1 \rightarrow f(g_1(s_1)) \rightarrow f(a_1)$
$f \circ g_1 \rightarrow f(g_2(s_2)) \rightarrow f(a_2)$
I thought that from here I could show that $f \circ g_1$ and $f \circ g_2$, and then show that $g_1$ and $g2$ are therefore equivalent, but nothing is clicking and I don't think I am on the right track. Am I going in the right direction?
HINT: The simplest approach here is to prove the contrapositive: show that if $g_1\ne g_2$, then $f\circ g_1\ne f\circ g_2$. This is logically equivalent to the desired implication.
If $g_1\ne g_2$, there is some $s\in S$ such that $g_1(s)\ne g_2(s)$. Now use the fact that $f$ is $1$-$1$ to tell you something about $(f\circ g_1)(s)$ and $(f\circ g_2)(s)$.