# If $f \circ g$ then $f$ is also injective…

Here I have such a problem:

We have two functions: $f:S \to T$ and $g:T \to U$.

Demonstrate that if $f \circ g$ is an injective function then $f$ is also an injective function.

Can you give me, please, a useful hint ? Thank you very much in advance.

• do you mean $g\circ f$? – Thomas Jan 8 '15 at 20:13
• One way: Start with: pick x and y in S such that f(x) = f(y). – Alex Zorn Jan 8 '15 at 20:18

I think it should be $g\circ f$, so the composition would be defined ($(g\circ f)(x)=g(f(x))$).
Hint: A function $f:A\to B$ is injective if and only if it is invertible from left, i.e. there exists $g:B\to A$ such that $g\circ f=I$ ($I$ is the identity map).