I have seen other questions and answers regarding this proof, but I am having trouble with the overall logic in it. Here is how the proof seems to look:
$\;$ $$\;\;\text{Suppose that g is not one-to-one. Then we can find distinct}\;x_1,x_2 \in X \; \text{for which}\; \:\:\:\:\:\;\;\;\;\;\;\;\;\;\;\;\;\; g(x_1)=g(x_2)=y. \;\text{But then }f \circ g(x_1)=f(y)=f \circ g(x_2) \;\text{so} \; f \circ g \;\text{is also not one-to-one. Therefore,}\;f\circ g \;\text{is not one-to-one unless g is also one-to-one. } $$
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Here is my reasoning which goes differently than this:
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$\;$ $\;\;\text{Suppose that g is not one-to-one. Then we can find distinct}\;x_1,x_2 \in X \; \text{for which}\; \:\:\:\:\:\;\;\;\;\;\;\;\;\;\;\;\;\; g(x_1)=g(x_2)=y. \;\text{Then it could be the case that we have} \; f \circ g(x_1),\;f \circ g(x_2), \; \text{where}\; g(x_1) =g(x_2)=y, \text{and therefore }f \circ g(x_1)=f \circ g(x_2)=f \circ g(y). \text{Since the outputs of}\;f \circ g(x_1),\;f \circ g(x_2) \;\text{are only equal when }g(x_1)=g(x_2), f \circ g \;\text{is one-to-one. But in this case, g does not need to be one-to-one in order for this to be true. } $
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If someone would be able to explain the error in my reasoning I would appreciate it! Thanks!
The problem with your reasoning is the following:
In the beginning you assume that $g(x_1) = g(x_2) = y$. Later you say that $f(g(x_1))$ is only equal to $f(g(x_2))$ if $g(x_1) = g(x_2)$. But you assumed that $g(x_1) = g(x_2)$ and thus it must be the case that $f(g(x_1)) = f(g(x_2))$. So f composed with g is not one-to-one if you assume that $g(x_1) = g(x_2)$.
This is an indirect proof of the lemma, meaning that you prove that if $g$ is not one-to-one, then $f \circ g$ is not one to one. This is equivalent to saying that if $f \circ g$ is one-to-one, then $g$ must be one-to-one.