I am planning on taking Abstract Algebra this upcoming semester, and I wanted to read up somewhat ahead of time. Unfortunately, it seems that I am getting stuck on what should be a rather elementary proof regarding composition of functions and how they are associative. I am using the Judson textbook, and it gives the following definition and proof.
Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be mappings. Define a new map, the composition of f and g from A to C, by $(g \circ f)(x) = g(f(x))$.
Then it says "Prove that $h \circ (g \circ f) = (h \circ g) \circ f$.
It then gives the first line of the proof as $(h \circ (g \circ f))(a) = h(g \circ f)(a)$. Now the reason that this step is confusing me is because it seems that the proof is operating under the assumption that $g \circ f$ can be written as its own mapping entirely, which I agree with. However, the definition provided only gives how to transform two individual functions $(g \circ f)(x) = g(f(x))$ with an input parameter. It does not give a definition as to how to transform $(g \circ f)$ into a map. Now, intuitively, it seems to me that $(g \circ f) = (g \circ f)(x)$, however, if we were to apply this definition to the proof, it would then become $(h \circ (g \circ f))(a) = (h \circ (g \circ f)(a))(a)$, which doesn't make sense to me.
My question is therefore, how would I be able to continue with the proof if I wanted to be able to show that $g \circ f$ is also a function?
The way I wanted to format the proof was as follows:
- Let $(h \circ (g \circ f))(a)$.
- By definition of ..., $(g \circ f)$ can be written as some function $j$.
- Therefore, $(h \circ (g \circ f))(a) = (h \circ j)(a)$.
- By definition of composition of functions, $(h \circ j)(a) = h(j(a))$.
- However, $h(j(a)) = h((g \circ f)(a))$ by construction.
- By definition of composition of functions, $h((g \circ f)(a)) = h(g(f(a))$
Thanks so much! I really appreciate it.