Thread: If gof is injective, then f is injective

1. If gof is injective, then f is injective

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

Show that if $g \circ f$ is injective, then $f$ is injective.

Here is what I did.

$Proof$. Spse. $g \circ f$ is injective and $f$ is not injective. Then $\exists x_1,x_2 \in A \ni f(x_1)=f(x_2)$ but $x_1 \neq x_2$.
But then $g \circ f(x_1)=g \circ f(x_2)$ would imply $x_1 \neq x_2$ thus $g \circ f$ is not injective. A contradiction.
Therefore if $g \circ f$ is injective, then $f$ is injective.

Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

Thanks

dubito

2. Originally Posted by dubito
Let $f:A \rightarrow B$ and $g: B \rightarrow C$ be functions.

Show that if $g \circ f$ is injective, then $f$ is injective.

Here is what I did.

$Proof$. Spse. $g \circ f$ and $f$ is not injective. Then $\exists x_1,x_2 \in f \ni f(x_1)=f(x_2)$ but $x_1 \neq x_2$.
But then $x_1,x_2 \in g \circ f$ would imply $x_1 \neq x_2$ thus $g \circ f$ is not injective. A contradiction.
Therefore if $g \circ f$ is injective, then $f$ is injective.

Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

Thanks

dubito
Seems you have confused yourself.
$
\exists x_1,x_2 \in f \ni f(x_1)=f(x_2)
$

etc make no sense to me.

gof is not a 'set', so how can you say x1,x2 belong to gof? What does it mean??

3. Originally Posted by dubito
Let $f:A \rightarrow B$ and $g: B \rightarrow C$ be functions.

Show that if $g \circ f$ is injective, then $f$ is injective.

Here is what I did.

$Proof$. Spse. $g \circ f$ and $f$ is not injective. Then $\exists x_1,x_2 \in f \ni f(x_1)=f(x_2)$ but $x_1 \neq x_2$.
But then $x_1,x_2 \in g \circ f$ would imply $x_1 \neq x_2$ thus $g \circ f$ is not injective. A contradiction.
Therefore if $g \circ f$ is injective, then $f$ is injective.

Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

Thanks

dubito
Assume that $f$ is not injective; that is, $\exists~x\neq y$ such that $f(x)=f(y)$. Then $g(f(x))=g(f(y))$, but since $x\neq y$, this implies that $g\circ f$ is not injective either. $\Rightarrow\Leftarrow$

4. Thanks guys! I reread my proof and I am like . @ aman_cc, I meant $x_1,x_2 \in A$ sry about that (if I remember correctly though I think my teacher used to do the same thing). @ redsoxfan325 .

dubito

5. Originally Posted by dubito
Thanks guys! I reread my proof and I am like . @ aman_cc, I meant $x_1,x_2 \in A$ sry about that (if I remember correctly though I think my teacher used to do the same thing). @ redsoxfan325 .

dubito
Another quick way

If f(x1)= f(x2) =>
gof(x1) = g(f(x1))=g(f(x2)) = gof(x2) =>
x1 = x2 (as gof is injective)

Hence done