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Is the inclusion function $ i_{AB}:A \to B$ the same function as the composition of the two identity functions $id_A:A \to A$ and $id_B:B \to B$ ? In other words, is the following equation true?

$$ i_{AB}=id_B \circ id_A $$

The answer seems to depend on the rules for composing functions. If, when composing $g \circ f$, the codomain of $f$ is allowed to be a subset of the domain of $g$, then the answer is yes. But if the codomain of $f$ is required to equal the domain of $g$, then the answer is no. So, which is it?

This particular "rule of function composition" has been discussed before. There was no consensus and it seemed inconsequential.
Composition of functions in Munkres' Topology
domain of composite function (is there a set rule)

But this is a case where it actually matters. If the answer to my question is yes, then it gives me a neat way of proving something is an inclusion function.

If the answer is no, then $id_B \circ id_A$ would be an illegal composition. In order to compose $id_A$ and $id_B$, you'd need to stick $i_{AB}$ in between them.

$$id_B \circ i_{AB} \circ id_A$$

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    On the other hand, am I being needlessly formal? – 2017-01-14

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Which conditions to use to define function composition is ultimately a convention, and which to choose depends on how you intend to manipulate function composition.

In the present case, if we write $i_{AB} = i_B \circ i_A$, then the result of composing the 2 bijections $i_A$ and $i_B$ is the injection $i_{AB}$, which is not a bijection unless $A = B$. So the formula is wrong, either because we refuse to consider $id_B \circ id_A$, or because considering it breaks theorems we wish to use afterwards.

We can still write $id_{AB} \mid^A = id_A$ to restrict the range $B$ of $i_{AB}$ to the image $A$ of $id_A$. The formula $id_B \circ i_{AB} \circ id_A$ is also well-formed.

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    Thanks. Yes, it's all conventions. But it's worthy trying to get all the conventions playing well together. So, allowing $i_{AB} = i_B \circ i_A$ breaks the rule that – 2017-01-16
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    Thanks. Yes, it's conventions all the way down. But it's worthy to get all the conventions playing well together so that, as you say, handy theorems don't break. So excellent answer. Allowing $i_{AB} = i_B \circ i_A$ breaks the rule that the composition of bijections should be a bijection. That's just the kind of thing I was worried about. This means then that we need to enforce the rule that when composing $g \circ f$ the range of $f$ must equal the domain of $g$ not just a subset. – 2017-01-16
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    Furthermore, we can say that the inclusion is a "prime function", that can't be factored. Which robs me of my clever way to show something is an inclusion. – 2017-01-16
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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. – 2017-01-23