Restrictions of compositions

Question

Considering two functions, $f,g:X\to X$ s.t. $Y\subseteq X$ is invariant for both $f$ and $g$, that is $f(Y)\subseteq Y$ and similar for $g$. Do we then have $f|_Y\circ g|_Y=(f\circ g)|_Y$. It seems trivial, but I'm not sure how to show it formally. It would be nice to see a set theoretic argument via the restrictions of relations.

Answer 1

The composition $f\mid_Y\circ g\mid_Y$ is well defined if and only if $g(Y)\subset Y$. If that is given, the equation is true as you can see below.

By definition of the restriction, you get $f(x)=f\mid_Y(x)$, $g(x)=g\mid_Y(x)$ and $f\circ g(x)=(f\circ g)\mid_Y(x)$ for all $x\in Y$.

For all $x\in Y$ this yields $$ (f\mid_Y\circ g\mid_Y)(x)=f\mid_Y(g\mid_Y(x))=f\mid_Y(g(x))=f(g(x))=f\circ g(x)=(f\circ g)\mid_Y(x). $$ (At the third equation you use $g(x)\in Y$.) Hence, $f\mid_Y\circ g\mid_Y=(f\circ g)\mid_Y$ holds.

Answer 2

It does look trivial: $\text {lhs}=f(g(y))=\text{rhs } \forall y\in Y$.

That is, we have $\bigcup_{y\in Y}\{(y,f(g(y)))\}$ for both functions.

Note: $f\mid _Y=f\circ i:Y\rightarrow Y$, where $i:Y\hookrightarrow X$ is the inclusion of $Y$ into $X$. I have written $Y$ for the codomain by invariance...