So, I have the statement that if $g: [k]\to[k]$ is a bijection, then $f: [n] \to[k]$ and $g\circ f: [n] \to[k]$ give the same partition of $[n].$
My thought process was that $g\circ f$ would be actually $[n] \to[k]\to[k]$ and because $[k] \to[k]$ in $[n] \to[k] \to[k]$ is a bijection, which is function $g,$ is a bijection, the partitioning of $[n]$ actually should be the same. I tried to use Bell Numbers to show that they have the same partition but it didn't work for me. I'm right now struggling how to prove this statement by inclusion-exclusion, induction or combinatorial proof, could anyone help me to how to prove this statement in either ways?
Partition as in quotient? Quotient by kernel? Sure, we have for all $a,b\in \{1,\ldots, n\}$ $$ f(a) = f(b) \Leftrightarrow gf(a) = gf(b)\qquad (g\text{ is injective}) $$ In other words $\ker f = \ker gf$.
The kernel of a map $h:[n]\to [k]$ is an equivalence relation (reflexive, transitive and symmetric) on $[n]$ defined by $$ (a,b)\in \ker h \subseteq [n]\times [n] \Leftrightarrow h(a) = h(b) $$ A set can be quotiented by an equivalence relation. The quotient is a set that consists of equivalence classes: $$ [n] / \ker h = \left\{ \overline{a} \mid a\in [n] \right \} $$ where $\overline{a} = \{x\in [n] \mid h(a) = h(x)\}$. This division into equivalence classes is also called a partition on $[n]$.