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Prove or disprove. Given any set X and given any functions f : X → X, g : X → X, h : X → X, if h is one-to-one and h o f = h o g, then f = g.

Let x belong to X

We know that (h o f)(x) = (h o g)(x)

(h o f)(x) = (h o g)(x) —> h(f(x)) = h(g(x))

Since h is one-to-one, we can say that f(x) = g(x)

Not sure where to go from this point...

Hello
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  • What is your definition of $f=g$? Isn't it $f(x)=g(x)$ for all $x\in X$? – Mohan May 09 '17 at 01:11
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    Correct f(x) = g(x) for all x in X. What are you asking? – Hello May 09 '17 at 01:24
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    Not sure wher to go? Well, since you finished the problem fairly quickly, I'd suggest a nice dinner and maybe a movie? You certainly don't have any more of this problem to do. – fleablood May 09 '17 at 01:28

2 Answers2

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You arrive to the point $f(x) = g(x)$ for all $x \in X$. This is your definition of $f = g$.

Tristan Batchler
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Even though $h$ is not assumed to be bijective, $f$ and $g$ must agree for all $x$ in $X$.

Proof:

Suppose there is an $x_0$ where they don't agree. Then

$f(x_0) \ne g(x_0)$

Since $h$ is 1:1,

$h(f(x_0)) \ne h(g(x_0))$

But this contradicts our assumption that $hf = hg$.

CopyPasteIt
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