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I have these two exercises for my math-study, and I don't really know how to prove them. Can you help me out?

A) Let f: X $\to$ Y and g: Y $\to$ Z be functions. Show that if g $\circ$ f is injective, then f must be injective.

B) Is it true that g must also be injective?

What I tried by A) so far was this:

Let x and x' be on X. If (g $\circ$ f)(x) = (g $\circ$ f)(x'), then x = x'

And by B) I think the answer is "no", but I can't explain why.

I know this is not much, but it's hard for me because I'm jus a beginner in math. Thanks in advance!

Martin Sleziak
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Peter
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  • Related questions: http://math.stackexchange.com/questions/63552/composite-functions-and-one-to-one, http://math.stackexchange.com/questions/229065/if-f-circ-g-is-onto-then-f-is-onto-and-if-f-circ-g-is-one-to-one-then-g, http://math.stackexchange.com/questions/83622/does-f-circ-g-injective-imply-f-injective-for-functions-f-ga-to-a and http://math.stackexchange.com/questions/22572/injective-and-surjective-functions – Martin Sleziak Feb 08 '15 at 19:15

1 Answers1

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For A), suppose by contradiction $f$ is not injective, i.e. there exist $x\neq x^\prime$ such that $f(x)=f(x^\prime)$. What can you say about $g\circ f(x)$ and $g\circ f(x^\prime)$ (and why does it contradict the fact that $g\circ f$ is injective?).

For B), indeed the answer is no, and one reason can be the fact that $f$ can ''hit'' only one portion of the domain of $g$. Consider the identity function $f\colon x\in [0,1] \mapsto x\in[0,1]$, and $g\colon x\in\mathbb{R}\mapsto x^2$. $g$ is not injective, and yet $g\circ f$ is (can you see why?).

Clement C.
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  • Thank you very much! I completely understand A) now. But I do not exactly get what you mean for B)? – Peter Feb 08 '15 at 18:49
  • $g$ is not injective on $\mathbb{R}$, yet $g\circ f$ is (on its domain, $[0,1]$). The reason being, the part of the domain on which $g$ is not injective is not reached ($g$, restricted to the non-negative reals, is injective; and $f$ only takes values in the non-negative reals). – Clement C. Feb 08 '15 at 18:56
  • But how can you speak of g∘f if the range of f is not the same as the domain of g? – Peter Feb 08 '15 at 19:16
  • It only needs to be a subset of the domain of $g$ to be well-defined. – Clement C. Feb 08 '15 at 19:30
  • I still don't see why g is not injective here. f(x) can only be 1 or zero, and therefore g(f(x)) can also be only 1 or 0. But anyway, thank you very much for your help – Peter Feb 08 '15 at 20:24
  • The interval $[0,1]$, not the set ${0,1}$. $f$ is the identity function that takes $0\leq x\leq 1$, and returns $x$. $g$ is the square function on the whole set of real numbers, which is not injective since $g(-1)=g(1)$ (for instance). – Clement C. Feb 08 '15 at 20:25
  • But -1 is not on the interval [0,1] – Peter Feb 08 '15 at 20:30
  • Indeed. $g$ by itself is a function from $\mathbb{R}$ to $\mathbb{R}+$, and is not injective. But $g\circ f$ is from $[0,1]$ to $\mathbb{R}+$ (it is the restriction of $g$ to $[0,1]$), and is injective. – Clement C. Feb 08 '15 at 20:31
  • oh yes I see it. Thanks! – Peter Feb 08 '15 at 20:32