What is wrong with the following argument about splitting an exact sequence of abelian groups?

Question

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).

Answer 1

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.