Imposter

Let $S = \{ 1, 2, 3 \}$ and $T = \{4, 5, 6, 7 \}$ . We know $g \circ f$ is injective by definition, so $f$ must be injective and $g$ must be surjective. However, $g$ doesn't have to be injective.

We can say that $f(1) = 4, f(2) = 5, f(3) = 6$ and $g(4) = 1, g(5) = 2, g(6) = 3$ . As a result, $g \circ f = \text{id}_{S}$ because the domain of $f$ does not give us an input with the image $7$ . Therefore, when we do $g$ of output of $f$ , there's no way to "not get an output". However, if we were to do $f \circ g$ , consider $g(7)$ , which has no output. $f$ has no way to deal with $g (7)$ (because what is $f$ of something that is undefined? Undefined.). Therefore, $f \circ g$ does not give $\text{id}_T$ . Therefore, there do exist functions $f$ and $g$ for which the question statement is true.