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Theorem. Let $f: W \rightarrow X, g : X \rightarrow Y, h: Y \rightarrow Z$ be maps. Then

(1) $h \circ ( g \circ f) = (h \circ g) \circ f$, i.e., the composition of maps is associative.

Proof is left as an exercise.

My attempt:

Proof: By definition of map composition

$h \circ ( g \circ f) = (h \circ g) \circ f$

$h \circ (W \rightarrow Y) = (X \rightarrow Z) \circ f $

$W \rightarrow Y \rightarrow Z = W \rightarrow X \rightarrow Z$

$W \rightarrow Z = W \rightarrow Z$

$\blacksquare$

I'd like to know, if either the proof and the way of writing it are correct. Or, if there's a more elegant way to write the proof (if correct).

Liesen, J., Mehrmann, V. 2015. Linear Algebra. Berlin, Germany.: Springer.

sputnik
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    Side note: I don't have someone (e.g. a teacher) to hand this exercises to, that's why I post the problems here. – sputnik May 17 '20 at 02:54
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    This doesn't prove associativity, you've just highlighted that these compositions are maps $W\to Z$. – Dave May 17 '20 at 03:24

1 Answers1

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To show equality $h\circ (g\circ f)=(h\circ g)\circ f$ we need to show that $(h\circ (g\circ f))(w)=((h\circ g)\circ f)(w)$ for each $w\in W$.

You should try this on your own, but below is a spoiler showing the argument for checking your solution.

Fix $w\in W$. By the definition of function composition, we have $$(h\circ(g\circ f))(w)=h((g\circ f)(w))=h(g(f(w)))$$ and on the other hand $$((h\circ g)\circ f)(w)=(h\circ g)(f(w))=h(g(f(w)))$$ so we're done.

Dave
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