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Old November 19th, 2009, 04:55 PM
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Exclamation Injective functions

Let f : A → B et g : B → C be 2 functions

Show that if g ◦ f is injective, then f has to be injective
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Old November 19th, 2009, 05:51 PM
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This isn't true, take \cdot ^2:\mathbb{R}\rightarrow[0,\infty) and \sqrt{}:[0,\infty)\rightarrow [0,\infty).
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Old November 19th, 2009, 06:05 PM
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Quote:
Originally Posted by Focus View Post
This isn't true, take \cdot ^2:\mathbb{R}\rightarrow[0,\infty) and \sqrt{}:[0,\infty)\rightarrow [0,\infty).
Notice that \sqrt{} \circ \cdot^2 is not injective.

As for the problem assume f(x)=f(y) then g(f(x))=g(f(y))which implies x=y by the injectivity of g\circ f
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