Given functions such that $f = \tilde{f}$ and $g = \tilde{g}$, then $g\circ f = \tilde{g}\circ\tilde{f}$.

Question

Verify the substitution property: if $f,\tilde{f}:X\rightarrow Y$ and $g,\tilde{g}:Y\rightarrow Z$ are functions such that $f = \tilde{f}$ and $g = \tilde{g}$, then $g\circ f = \tilde{g}\circ\tilde{f}$.

MY ATTEMPT

We say two functions $f,g:X\rightarrow Y$ are equal iff $f(x) = g(x)$, for all $x\in X$. Hence, for arbitrary values of $x\in X$, we have that $$\begin{align} (g\circ f)(x) &= g(f(x))\\ & = g(u) \\ &= \tilde{g}(u)\\ & = \tilde{g}(f(x)) \\ &= \tilde{g}(\tilde{f}(x)) \\ &= (\tilde{g}\circ\tilde{f})(x), \end{align}$$ which proves that $g\circ f = \tilde{g}\circ\tilde{f}$.

Can someone check if I am reasoning rightly?

Answer 1

Your proof is fine.

However, be careful: here $g:Y\to Z$ is already given. It would help, instead, to write

$h,k:A\to B$ are equal iff $h(a)=k(a)$ for all $a\in A$.

Stating that $u=f(x)$ is prudent too. In fact, you could do away with any such substitution, like so:

$$\begin{align} (g\circ f)(x) &= g(f(x))\\ & = g(\tilde{f}(x)) \\ &= \tilde{g}(\tilde{f}(x)) \\ &= (\tilde{g}\circ\tilde{f})(x). \end{align}$$