Is it true if $g\circ f$ is injective then $g$ is injective?

Question

I need to know if this statement is true or false:

If $g\circ f$ is injective then $g$ is injective.

I couldn't try to prove this statement. I was thinking to show a counter-example: $$\begin{array}{lll} f\colon\{1\}\to\{2,3\},&f(1)=1&\\g\colon\{2,3\}\to\{1\},&g(2)=1,&g(3)=1 \end{array}$$ But then I can't seem to show that $g\circ f$ is injective.

I would really appreciate any help on this. Thanks in advance.

The correct statement is that if $g \circ f$ is injective then $\left . g \right |_{f(A)}$ is injective. — Ian, Nov 09 '14 at 18:24
It only took four years, but finally I spot a student of mine asking here about homework! And you'd think it would happen sooner with my penchant for sophisticated questions. Congrats, you're the first person I notice asking here for help. — Asaf Karagila, Nov 10 '14 at 21:44

score 2 · Accepted Answer · answered Nov 09 '14 at 18:15

2

(I assume that you mean $f(1)=2$, because $f(1)=1$ is not possible since $f\colon\{1\}\rightarrow\{2,3\}$).

Note that $g\circ f$ has domain $\{1\}$. Since the domain has only one element, $g\circ f$ is necessarily injective. Indeed, there does not exist $x,y$, different elements such that $f(x)=f(y)$. Thus the statement

If $g\circ f$ is injective then $g$ is injective.

is false.

answered Nov 09 '14 at 18:15

zarathustra

4,891

Alright so I need to prove that? – matan Nov 09 '14 at 18:17
There is nothing to add to the proof! What I wrote is a proof of the fact that $g\circ f$ is injective. Since $g$ is not injective, it follows that the statement you had to discuss is false. – zarathustra Nov 09 '14 at 18:18
As domain is singleton then map has to be injective.From definition of mapping.Nothing to add in the srgument given by @rosebud – Ripan Saha Nov 09 '14 at 18:18
Ok I understand now – matan Nov 09 '14 at 18:21

Is it true if $g\circ f$ is injective then $g$ is injective?

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