For I'm learning Category Theory, sometimes I encounter people treating function composition like if it's equal to another single function (i.e. not just results the same!). As an example, below I quoted the proof from Conceptual Mathematics: A first introduction to Categories.
Note: authors does not just equate result of $x$ to $s∘y$, but rather say that $x$ is $s∘y$, that's the gist of the proof. But the assumption would wreak havoc onto commutative diagrams! I.e. if they're the same arrows, $T = B$ must be the same objects too ($y$ becomes identity, $x$ must be $s$); and we could any commutative diagram collapse to just a pair of objects, killing out e.g. products.
I'd think authors misunderstand something, but I encountered alike treatment before in other sources, so perhaps it's mine incomprehension. What do I miss?
Proposition 1: If a map $A \xrightarrow{f}B$ has a section, then for any $T$ and for any map $T \xrightarrow{y} B$ there exists a map $T \xrightarrow{x}A$ for which $f ∘ x = y$.
Proof: The assumption means that we have a map $s$ for which $f ∘ s = 1_B$. Thus for any given map $y$ as below
$\hskip3in$
we see that we could define a map $x$ with at least the correct domain and codomain by taking the composite $s$ following $y$
$$x=s∘y$$
Does this map x actually satisfy the required equation? Calculating
$$f ∘ x = f ∘ (s∘y) = (f∘s)∘y=1_B ∘ y = y$$
we see that it does.
Yes, this is precisely how composition of morphisms works.
No. $s \circ y$ is a morphism from $T$ to $A$. $x$ is also a morphism from $T$ to $A$. So everything is fine.