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Show that $g$ is injective if and only if for any set $A$ and pair of maps $f_1: A \rightarrow X$ and $f_2: A \rightarrow X,$ we have $g\circ f_1=g \circ f_2$ implies $f_1 = f_2.$

I can do the forward direction:

Proof: $(\Rightarrow)$ Assume $g$ is injective. Then, for all $x_1,x_2 \in X$, $g(x_1) = g(x_2)$. Let $f_1, f_2$ be maps $f_1: A \rightarrow X$ and $f_2: A \rightarrow X,$ for some $A$.

Assume that $g\circ f_1=g \circ f_2$. Then, $(g\circ f_1)(a)=(g \circ f_2)(a)$, for all $a \in A$.

So, $g\circ (f_1(a))=g \circ (f_2(a))$. Since $g$ is injective, $f_1 (a) = f_2 (a)$ for all $a \in A$. Then, $f_1 = f_2$.

I have no idea how to do the other direction, can someone give me a hint?

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2 Answers 2

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Take $$f_1=f_2=\operatorname{Id}_{X}.$$

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  • $\begingroup$ so $g(x_1)=g(x_2) \Rightarrow g(Id_X(x_1)) = g(Id_X(x_2) ) \Rightarrow Id_X(x_1) = Id_X(x_2) \Rightarrow x_1 = x_2$ $\endgroup$
    – user35687
    Commented Jan 31, 2020 at 15:29
  • $\begingroup$ @user35687 Yes, that's correct. $\endgroup$
    – cqfd
    Commented Jan 31, 2020 at 15:30
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Suppose $g$ is not injective and $g(x_1)=g(x_2)=y$ where $x_1\ne x_2$. Now let $f_1(a_1)=x_1$, $f_1(a_2)=x_2$, $f_2(a_1)=x_2$, $f_2(a_2)=x_1$, all other mappings of the functions being equal. Then the implication in the latter part of the problem is not satisfied, since $f_1\ne f_2$ yet $g\circ f_1=g\circ f_2$.

Since we have shown the converse of the forward direction, the proof is complete.

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