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In Munkres' book "Topology", he writes that:

Given functions $f:A\to B$ and $g:B \to C$, .... the composition $g \circ f $ is defined only when the range of $f$ equals the domain of $g$.

But, isn't it enough that the range of $f$ is a subset of the domain of $g$?

Why does he say that the range of $f$ needs to be equal to the domain of $g$?

user74261
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  • Subset seems fine to me. – copper.hat May 19 '15 at 16:58
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    By range Munkres probably means codomain rather than image. –  May 19 '15 at 16:59
  • @Bye_World Even if he means co-domain, why does the codomain need to be equal? – user74261 May 19 '15 at 17:02
  • To make sure you don't try to plug in things outside of the domain of $g$. Remember if the image of $f$ is a subset of the domain of $g$, you can just define the codomain to be equal to the domain of $g$, so it doesn't really hinder you at all. –  May 19 '15 at 17:03
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    If the codomain weren't equal, you'd have to lug around the assumption that the image of $f$ is in the domain of $g$. It's much neater by just saying the codomain is the domain of $g$. – GPerez May 19 '15 at 17:04
  • I think it was answered with the typo fix! – Jesse P Francis May 19 '15 at 17:07

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Here's one good reason why Munkres requires that when composing $g \circ f$, the range of $f$ must be equal to domain of $g$ and not just a subset. If we allowed it otherwise, it would break the theorem that says the composition of bijections is a bijection.

TJCrow
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  • Oops, I mis-stated the rule. It should be that: When composing $g \circ f$, the codomain of $f$ must equal the domain of $g$. This keeps these 3 theorems in tact: (1) The composition of injective functions is injective, (2) The composition of surjective functions is surjective, and (3) The composition of bijective functions is bijective. – TJCrow Jan 23 '17 at 18:29