Disprove: $g \circ f$ is injective $ \implies g$ is injective

164 Views Asked by At

Can someone verify the work I have done for this proof? Thanks in advance.

Let $f:A\rightarrow B$ and $g:B \rightarrow C$ be functions.


Suppose $A = \mathbb{N}_0, B = \mathbb{Z}$ and $C = \mathbb{N}_0$.

Where $\mathbb{N}_0 = \mathbb{N} \cup \{0\}$.

Consider the function $g: \mathbb{Z} \rightarrow \mathbb{N}_0$

$g(x) = x^2$.

$g$ is not injective from $\mathbb{Z} \rightarrow \mathbb{N}_0$, as seen by

$g(2) = g(-2) = 4$.

Consider the function $f:\mathbb{N}_0 \rightarrow \mathbb{Z}$

$f(x) = x$, such that

$(g\circ f)(x) = g(f(x)) = x^2$.

This function is injective from $\mathbb{N}_0 \rightarrow \mathbb{N}_0$.

Thus $g \circ f$ is injective $\land$ $g$ is not injective, so we have disproven the hypothesis.