As remarked in the "talk" part of the Wikipedia article, the proof is done with elements of a set and functions. I guess it's possible to carry it out purely with "objects" and "arrows"

Who volunteers to do that?

Edit: If possible without that Freyd embedding theorem mentioned in the talk.

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    $\begingroup$ Why don't you take a crack at it? $\endgroup$
    – Ian Coley
    Commented Apr 10, 2014 at 19:48
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    $\begingroup$ The obvious part is that a biproduct implies that the short exact sequence splits on the right and on the left. For the converse, the fact that the short exact sequence splits on either side with give a diagram that looks pretty much like either a product or a coproduct, here is where I am $\endgroup$
    – Noix07
    Commented Apr 10, 2014 at 19:54
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    $\begingroup$ This is proposition 1.8.7 in Borceux's Handbook of Categorical Algebra, Vol II - Categories and Structures. $\endgroup$
    – user153312
    Commented Jan 7, 2015 at 19:53

2 Answers 2


Notations: The short exact sequence splits on the left

$$ 0 \longrightarrow A \underset{\underset{\phi}{\longleftarrow}}{\overset{f}{\longrightarrow}} B \overset{g}{\longrightarrow} C \longrightarrow 0 \quad, \quad \phi\circ f =\mathrm{id}_A$$

Let's show that $B \cong A\oplus C$

$\underline{1)}\ $$\enspace \phi\circ f = \mathrm{id}_A\ \Rightarrow\ P:= f\circ \phi $ is an idempotent ($P\circ P=P$).

  • Any arrow in an abelian category has an image so that $P$ splits into $ B \overset{j_P}{\longrightarrow} \mathrm{Im}(P) \overset{i_P}{\longrightarrow} B $ with $j_P$ epi and $i_P$ mono (cf. the equivalent def. of image and the property that $j_P$ is epi in categories with finite limits and colimits)

  • $j_P$ is actually a left inverse of $i_P$: $$i_P\, j_P\,i_P\, j_P= P^2 = P = i_P\, j_P\quad\underset{i_p\ \text{mono}}{\overset{j_P\ \text{epi}}{\Longrightarrow}}\quad j_P\,i_P= \mathrm{id}_{\mathrm{Im}(P)}$$

  • $j_P\circ f$ is actually an isomorphism $A \cong \mathrm{Im}(P)\ $: $f\circ \phi$ is a factorization of $P$ so that by the universal property of the image (in the sense of "initial factorization") there exists a unique morphism $v: \mathrm{Im}(P)\to A$ such that $i_P =f\, v$ and $\phi =v\, j_P$ whence

    $$(j_P\, f)\, v = j_P\, i_P = \mathrm{id}_{\mathrm{Im}(P)}\quad ,\quad v\,(j_P\, f)= \phi \, f = \mathrm{id}_A$$

(This is Proposition 10.2 p.12 from "Theory of Categories", Barry Mitchell: In a balanced category, if a morphism has an image and an (Epi-Mono) factorization then the mono is the image)

$\underline{2)} $ Hom sets of an abelian category are abelian groups so that it makes sense to define $$Q := \mathrm{id}_B -P\quad \in\big(\mathrm{Hom}(B,B),+\big)$$

As previously, $Q$ factorizes as $Q=i_Q \circ j_Q$ and with $i_P, i_Q$ mono, $j_P, j_Q$ epi $$ QP = QP = 0\quad \Longrightarrow\quad j_P\circ i_Q = 0 \enspace ,\enspace j_Q \circ i_P = 0 \qquad (Eq)$$

$\underline{3)} $ The two maps $\ i_P: \mathrm{Im}(P)\rightarrow B\ ,\ i_Q: \mathrm{Im}(Q)\rightarrow B $ define by the universal property of the coproduct a map $\Phi: \mathrm{Im}(P)\coprod \mathrm{Im}(Q) \rightarrow B$ (this could also be written $i_P+i_Q$, + symbolically if we were not in an abelian category )

and the two maps $j_P : B \rightarrow \mathrm{Im}(P)\ ,\ j_Q : B \rightarrow \mathrm{Im}(Q) $ a map $\Psi: B\rightarrow \mathrm{Im}(P)\prod \mathrm{Im}(Q)$

Using $j_P\circ i_{P} = \mathrm{id}_{\mathrm{Im}(P)},\ j_Q\circ i_Q = \mathrm{id}_{\mathrm{Im}(Q)},\ (Eq)$ and the definition of biproduct, one checks the isomorphism $B\cong \mathrm{Im}(P)\oplus \mathrm{Im}(Q)$, i.e. $$ \Phi\circ \Psi =P+Q = \mathrm{id}_B\quad \text{and} \quad \Psi\circ \Phi =\mathrm{id}_{\mathrm{Im}(P)\oplus\mathrm{Im}(Q)}$$

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    $\begingroup$ Why does it suffice to prove: $B\cong A\oplus C$ $\endgroup$ Commented Jan 15, 2017 at 0:24
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    $\begingroup$ Actually, I wrote everything very roughly... I would now say that we want something slightly more, an isomorphism of two short exact sequences: $A\oplus C$ is both the product and coproduct of $A, C$ s.t. the canonical injections and projections satisfy $\pi_A\, \iota_A=\mathrm{id}_A,\ \pi_C\, \iota_C=\mathrm{id}_C,\ \pi_C\, \iota_A=\pi_A\, \iota_C$. This is expressed as a short exact sequence $0\to A\to A\oplus C\to C \to 0$ that splits. One also writes the original s.e.s. $0\to A\to B\to C \to 0$. A morphism of exact sequence is a triple $(A\to A, B\to A\oplus C, C\to C)$ s.t. the squares... $\endgroup$
    – Noix07
    Commented Feb 14, 2017 at 0:02
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    $\begingroup$ [...] commute. So instead of just having $B\cong A\oplus C$, an isomorphism of short exact sequences consists in three isomorphisms. Of course, one would usually chose $(A\to A, C\to C)$ to be the identity of the respective object but then $B\to A\oplus C$ needs to satisfy some conditions. Now, if our original s.e.s. is isomorphic to the second one which splits, then it also splits. (Sorry for the late answer, I guess you may have figured it out since ages...) $\endgroup$
    – Noix07
    Commented Feb 14, 2017 at 0:10
  • $\begingroup$ No worries, I will think about it after my exams! Thanks! :) $\endgroup$ Commented Feb 14, 2017 at 1:13

Just saw this: there's a far simpler and more intuitive proof using the short five lemma.

Letting $i:A\to A\oplus C$, $j:C\to A\oplus C$, $p:A\oplus C\to A$, $q:A\oplus C\to C$ be the $4$ morphisms characterizing $A\oplus C$, consider the following diagram

enter image description here

Clearly, the upper short exact sequence splits on the left. $\phi$ is defined as the unique morphism such that $t=p\circ\phi$, $g=q\circ\phi$. It's easy to show that this diagram commutes. Thus, by the short five lemma, $\phi$ is an isomorphism.

A pretty similar proof works for splitting on the right as well.

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    $\begingroup$ The snake lemma can be used as well. Note also that the proof for the right one follows from the left one, as the argument for the left one is valid for all abelian categories, and so one can reduce the right case to the left one by taking the dual of an abelian category. $\endgroup$
    – wlad
    Commented Mar 23, 2023 at 11:08
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    $\begingroup$ That is one beautiful diagram! :) $\endgroup$ Commented May 11, 2023 at 0:22

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