One of my favourite elementary results in group theory is Goursat's Lemma. This lemma characterises the subgroups of a direct product of groups in terms of fibred products.

Indeed, let $L$ and $R$ be groups and let $G < L \times R$ be a subgroup of their direct product. We have natural projections $\pi_L : L \times R \to L$ and $\pi_R: L \times R \to R$. We may assume without loss of generality that $\pi_L$ and $\pi_R$ are surjective when restricted to $G$. Let $L_0 = \pi_L(G \cap \ker\pi_R)$ and $R_0 = \pi_R(G \cap \ker\pi_L)$ denote, respectively, normal subgroups of $L$ and $R$. Goursat's Lemma is the observation that $G$ defines an isomorphism $L/L_0 \cong R/R_0$.

Proof: If $x \in L$, let $y \in R$ be such that $(x,y) \in G$. Such a $y$ exists because of surjectivity of $\pi_L : G \to L$, but it need not be unique. Nevertheless the coset $y R_0$ is well defined. This procedure then defines a homomorphism $L \to R/R_0$ whose kernel is precisely $L_0$ and which is surjective because $\pi_R : G \to R$ is.

If we let $\lambda: L/L_0 \to F$ and $\rho: R/R_0 \to F$ be isomorphisms to the same abstract group $F$, then we may identify $G$ with the fibred product $$ G = \lbrace (x,y) \in L \times R ~|~ \lambda(x L_0) = \rho(y R_0) \rbrace .$$

There are similar results for Lie subalgebras of a direct sum of Lie algebras, and probably also in other categories. This suggests the following categorical slogan:

"subobjects of a product are pullbacks"

(Well, at least subobjects with the property that the composition with the epis in the product are also epis.)

Of course, this is not going to be true in all categories, which prompts the following question.


Let $\mathcal{C}$ be a category and $L,R$ be objects whose product $L \times R$ exists. Let $G \to L \times R$ be a monomorphism such that the compositions $G \to L \times R \to L$ and $G \to L \times R \to R$ are epimorphisms.

What must we demand of $\mathcal{C}$ so that there exist epimorphisms $L \to F$ and $R \to F$ such that $$\begin{matrix} G & \rightarrow & L \cr \downarrow & & \downarrow \cr R & \rightarrow & F \end{matrix} $$ is a pullback?


One would be tempted to call these categories Goursat categories, but alas the name seems to be taken already for what seems like a different concept.

Thanks in advance.


To make life simple, suppose that finite limits and finite colimits exist. If we work with regular epimorphisms rather than epimorphisms then your condition is equivalent to saying that for any two (regular) epimorphisms $G\to L$ and $G\to R$, if you form the pushout $F$ then the canonical comparison from $G$ to the pullback $L\times_F R$ is a regular epimorphism.

This is true in any exact Mal'cev category: see Theorem 5.7 of

Carboni, Kelly and Pedicchio, Some remarks on Maltsev and Goursat categories, Applied Categorical Structures 1:385-421, 1993.

Here exact means that the category (i) has finite limits (ii) has regular epi-mono factorizations (iii) the pullback of a regular epi is a regular epi (iv) any equivalence relation is the kernel pair of some map (one can choose the map to be the coequalizer) and Mal'cev can be characterized in many ways. For example, it says that if R and S are equivalence relations on some object A, then RS=SR.

In fact if the category is regular, in the sense that (i)-(iii) hold, then your condition is equivalent to being exact and Mal'cev.

By the way, the Goursat categories you mention are only slightly weaker: they have RSR=SRS rather than RS=SR. You can still prove your condition for Goursat categories if you suppose that at least one of $G\to L$ and $G\to R$ is a split epimorphism (i.e. has a section), and in fact this can be used to characterize Goursat categories. See link text

  • $\begingroup$ Many thanks. This is a very helpful answer. $\endgroup$ Nov 20 '10 at 5:39
  • $\begingroup$ One has to wonder whether it is not a coincidence that Goursat's Lemma is often true in so-called Goursat categories... $\endgroup$ Nov 21 '10 at 8:43
  • $\begingroup$ Not entirely a coincidence, I'd say - probably two very similar results for groups have been taken as the heart of Goursat's result, but these two results turn out to have slightly different strengths. $\endgroup$
    – Steve Lack
    Nov 22 '10 at 1:59
  • 1
    $\begingroup$ The link is broken. $\endgroup$
    – arsmath
    Apr 14 '20 at 8:14

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