There exist partition of set $X$ due to relation $R$ and surjection $g: X\to X|_R$ and injection $h:X|_R \to Y$ such as $h \circ g=f$

Question

$f: X\to Y$ is function. Prove: There exist partition of set $X$ due to relation $R$ on $X$ and surjection $g: X\to X|_R$ and injection $h:X|_R \to Y$ such as $h \circ g=f$

Answer 1

Any relation $R$ generates an equivalence relation $E$ wich is by definition the intersection of all equivalence relations that contain $R$ as a subset.

Any equivalence relation $E$ on a set $X$ corresponds with a partition $P$ on $X$ that has the equivalence classes of $E$ as its elements.

Denoting the equivalence class represented by $x\in X$ as $[x]$ we have the natural map prescribed by $x\mapsto[x]$ wich is a surjection.

If $f:X\to Y$ is a function then we have the relation $R$ on $X$ prescribed by $aRb\iff f(a)=f(b)$. This relation is an equivalence relation allready, so $R=E$ here.

(I suspect that the relation $R$ mentioned in your question has this link with the function $f$)

If in that case $P$ denotes the partition then next to the natural map $X\to P$ we also have the map $P\to Y$ prescribed by $[x]\mapsto f(x)$. It is well defined and injective.

Denoting this map with $h$ and the natural map with $g$ we have $f=h\circ g$.