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On page 147 of Hatcher's Algebraic Topology, he states the Splitting Lemma, which says the following:

Lemma. For a short exact sequence of abelian groups $$\require{AMScd} \begin{CD} 0 @>>> A @>i>> B @>j>> C @>>> 0 \end{CD} $$ the following statements are equivalent.

  1. (a) There is a homomorphism $p: B\to A$ such that $pi = 1_A$
  2. (b) There is a homomorphism $s:C\to B$ such that $js = 1_C$.
  3. (c) There is an isomorphism $B\to A\oplus C$ making a commutative diagram as at the right (see below), where the maps on the bottom row are the obvious ones.

I am confused because I accidentally proved that (c) is true regardless of (a) or (b). In fact, since (c) implies (a) and (b), I have accidentally proven that (a) and (b) are true whenever there is an exact sequence of abelian groups, which is incorrect. Can someone please tell me what is wrong with my following argument? I believe I'm missing something extremely obvious.

Consider the short exact sequence of abelian groups (I can't draw diagonal arrows, so I resort to this rectangular diagram). $$\require{AMScd} \begin{CD} 0 @>>> A @>i>> B @>j>> C @>>> 0\\ @| @| @VVV @| @| \\ 0 @>>> A @>>> A\oplus C @>>> C @>>> 0 \end{CD} $$ We are given that the top row is exact, and the bottom row is clearly exact due to the maps $$ a\mapsto (a,0) \qquad \qquad (a,c)\mapsto c $$ Moreover, the four "equal" arrows are isomorphisms. Thus, by the (short) five lemma the middle arrow is an isomorphism, proving (c).

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    $\begingroup$ Why would an appropriate middle vertical arrow ($B \rightarrow A \oplus C$) exist in all cases ? How is it defined here ? $\endgroup$
    – FiMePr
    Commented Dec 16, 2021 at 17:11
  • $\begingroup$ So is the crux of the argument that (a) and (b) each allow for a construction of the map $B\to A\oplus C$, and it follows from the five lemma that any such map is an isomorphism? $\endgroup$
    – jsborne
    Commented Dec 16, 2021 at 17:15
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    $\begingroup$ On second thought, my previous comment cannot be correct. The trivial map $B\to A\oplus C$ sending everything to zero is not an isomorphism. $\endgroup$
    – jsborne
    Commented Dec 16, 2021 at 17:18
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    $\begingroup$ Would you kindly state the exact form of the Splitting Lemma in Hatcher as well as what (a), (b), and (c) are? I'd rather not have to go to the library to be able to answer the question. $\endgroup$ Commented Dec 16, 2021 at 17:25
  • $\begingroup$ I have made an edit to my question. Thanks in advance. $\endgroup$
    – jsborne
    Commented Dec 16, 2021 at 17:30

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Thank you @FiMePr for a hint that eventually cleared everything up. Indeed, generically a map $B\to A\oplus C$ doesn't exist. The propositions (a), (b) allow for the explicit construction of maps $B\to A\oplus C$ and $A\oplus C\to B$, respectively. It is then easy to check that these maps make the squares of which they are a part of commutative. One then invokes the five lemma to prove that the maps are in fact isomorphisms, completing the proof.

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  • $\begingroup$ In other words, you proved that if there is any arrow $B \to A \oplus C$ at all making the diagram commute, then the arrow is an isomorphism. But you didn't actually show that there is an arrow at all. $\endgroup$ Commented Dec 16, 2021 at 17:51

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