The statement that surjective maps are epimorphisms in the category of sets can be shown in a constructive way.

What about the inverse?

Is it possible to show that every epimorphism in the category of sets is surjective without reverting to a proof by contradiction / negation?

  • 2
    $\begingroup$ Given the epimorphism $f$ define $g:Y\to\lbrace 0,1\rbrace$ by $g(y)=1$ if $y\in f(X)$ and $g(y)=0$ else. For the constant function $h(y)=1$ you have $g\circ f=h\circ f$ so that $g=h$. Hence, every $y\in Y$ belongs to $f(X)$ and $g$ is surjective. $\endgroup$ Commented Aug 18, 2014 at 10:24
  • 6
    $\begingroup$ @JochenWengenroth: Why is $y\in f(X)$ decidable? $\endgroup$ Commented Aug 18, 2014 at 10:44
  • 2
    $\begingroup$ I think that depend on what you mean by "constructive". if it is "topos valid"/without excluded middle then it is indeed true : consider two maps from $Y$ to the object obtained from $Y$ be identifying any two points which are in the image of $X$. $\endgroup$ Commented Aug 18, 2014 at 11:22
  • 3
    $\begingroup$ @AndreasCaicedo, I do not understand why you have removed tag "set-theory". As for me, the question is exactly about set theory (moreover, it is formulated in such a way, that it is about the standard set theory). $\endgroup$ Commented Aug 18, 2014 at 19:50

4 Answers 4


Theorem: Every epi is surjective.

Proof. Let $h : A \to B$ be an epimorphism. We define maps $f, g : B \to \mathcal{P}(B)$ by \begin{align*} f(b) &= \{b\} \cap \mathrm{im}(h)\\ g(b) &= \{b\} \end{align*} where we recall that $\mathrm{im}(h) = \{b \in B \mid \exists a \in A \,.\, h(a) = b\}$.

For every $a \in A$ we have $f(h(a)) = \{h(a)\} = g(h(a))$, therefore $f = g$ as $h$ is epi. Now, for every $y \in B$ we have $\{y\} = g(y) = f(y) = \{y\} \cap \mathrm{im}(h)$, therefore $y \in \mathrm{im}(h)$. QED.

Supplemental 2022-01-09: Here is an improved version which uses a smaller codomain. We write $\Omega$ for the subobject classifier (the set of truth values).

Proof. Let $h : A \to B$ be an epimorphism and $b \in B$. We define maps $f, g : B \to \Omega$ by \begin{align*} f(b') &{}\mathbin{{:}{=}} (b = b' \land \exists a \in A . h(a) = b')\\ g(b') &{}\mathbin{{:}{=}} (b = b'). \end{align*}

For every $a \in A$ we have \begin{align*} f(h(a)) &\Leftrightarrow (b = h(a) \land \exists a' \in A . h(a') = h(a)) \\ &\Leftrightarrow (b = h(a)) \\ &\Leftrightarrow g(h(a)), \end{align*} therefore $f = g$ as $h$ is epi. Now \begin{align*} \top &\Leftrightarrow b = b \\ &\Leftrightarrow g(b) = f(b) \\ &\Leftrightarrow b = b \land \exists a \in A . h(a) = b \\ &\Leftrightarrow \exists a \in A . h(a) = b. \quad \Box \end{align*}

  • $\begingroup$ Is this really different from the argument in my comment? $\endgroup$ Commented Aug 19, 2014 at 7:43
  • 2
    $\begingroup$ It is not really different, but I would say that it is less mysterious. $\endgroup$ Commented Aug 19, 2014 at 8:31
  • $\begingroup$ Thanks for the many answers; I liked this one best: I'm actually not working in Set but rather in Enriched Categories so this should be a proof that might translate to my setting. $\endgroup$ Commented Aug 19, 2014 at 10:36
  • 1
    $\begingroup$ @GarlefWegart So what are you really trying to prove? That every epimorphism is a extremal epimorphism? Or regular? Or split? $\endgroup$
    – Zhen Lin
    Commented Aug 19, 2014 at 16:46
  • 9
    $\begingroup$ @JochenWengenroth This proof would be the same as the one in your comment if classical logic were available. As Emil pointed out in the comment after yours, your formulation assumes that every $y$ is either in $f(X)$ or not, so that the two cases in your definition of $g$ cover every $y$. Andrej's answer avoids that case distinction, so it is valid even in constructive logic. The price for that is that, unlike your $g$ whose values are 0 or 1, his $f$ takes values not (constructively) known to be $\varnothing$ or $\{b\}$. $\endgroup$ Commented Aug 20, 2014 at 3:02

The statement "a morphism is a surjection iff it is an epimorphism" holds in every topos, regardless of the law of excluded middle.

The precise proof depends on your notion of "surjection" (in a topos all reasonable internal notions of a surjection coincide --- in fact, due to the above statement, one may define a surjection as an epimorphism).

Perhaps the most obvious notion is: a morphism $s \colon A \rightarrow B$ is a surjection if whenever $b \in B$ then $\underset{a\in A}\exists s(a) = b$ in the internal logic of the category. If a category is regular, then such surjections coincide with covers. And covers are another obvious notion for surjections: a morphism $s \colon A \rightarrow B$ is a surjection (i.e. cover) if in the image-factorisation $A \rightarrow s[A] \rightarrow B$ the monomorphism $s[A] \rightarrow B$ is iso.

Since every topos is a balanced category, in every topos covers coincide with epimorphisms.


Here's a proof in constructive set theory (probably just a rephrasing of the topos theoretic proof but you might find it useful).

Let $h : A \twoheadrightarrow B$ be an epimorphism. Define $$ C := \{\{0\}\} \cup \bigcup_{b \in B}\{\{x \in \{0\} \;|\; \exists a \in A\; h(a)=b\}\} $$ (If the powerset axiom is available, one can alternatively use $C := \mathcal{P}(\{0\})$)

Define functions $f, g : B \rightarrow C$ as follows. $$ f(b) := \{ x \in \{0\} \;|\; \exists a \in A\;h(a) = b \} \\ g(b) := \{0\} $$

We clearly have $f \circ h = g \circ h$, so since $h$ is an epimorphism, we get $f = g$. Now for any $b \in B$, we have that $f(b) = g(b)$. Therefore the set $\{x \in \{0\} \;|\; \exists a \in A \; h(a) = b \}$ is inhabited, and so there exists $a$ in $A$ such that $h(a) = b$.

  • 1
    $\begingroup$ Actually, you need the power-set axiom to write your first definition. The second one, just says that $C := \mathcal{P}(\{0\}) = \Omega$, where $\Omega$ is the set of internal truth-values. $\endgroup$ Commented Aug 18, 2014 at 13:49
  • 1
    $\begingroup$ Then, $f, g : B \rightarrow \Omega$ become predicates on $B$, where $f$ expresses the statement that $h$ is internally surjective, and $g := \top_B$. Therefore, $f = g$ says that "$h$ is internally injective" is true. $\endgroup$ Commented Aug 18, 2014 at 13:55
  • $\begingroup$ I claim that the first definition of C requires only bounded separation, strong collection and union (and pairing), so it's provably a set in CZF, for instance. I maybe should have been clearer that the first definition of C gives a different, in general strictly smaller set than the second. $\endgroup$
    – aws
    Commented Aug 18, 2014 at 14:48
  • $\begingroup$ Sorry, then I still don't understand your construction --- could you write it down in a bit more formal way? I don't claim that you're wrong, I just want to understand what exactly you are doing. I think your proof cannot be carried to a general predicative universe --- I claim that in any regular category the statement "epi iff surjection" is equivalent to "bimorphism iff iso" (i.e. to the statement that the category is balanced). Clearly, every topos is balanced (because $\Omega$ classifies morphism), (cont...) $\endgroup$ Commented Aug 18, 2014 at 17:47
  • 1
    $\begingroup$ @MichalR.Przybylek Pretoposes are balanced. Do you mean quasitoposes? $\endgroup$
    – Zhen Lin
    Commented Aug 18, 2014 at 19:25

I want to write down a proof that comes naturally, in a way. This proof assumes that you can form the quotient of a set modulo an equivalence relation, but does not require powersets. (So it works in any pretopos.)

We have sets $ A $ and $ B $ and a function $ f $ from $ A $ to $ B $. We know that $ f $ is an epimorphism in the category of sets, and we want to prove that $ f $ is a surjection.

Consider the cokernel pair of $ f $, that is the pushout of $ f $ along itself. Just as epimorphisms in an abelian category have trivial cokernels, so epimorphisms in more general categories have trivial cokernel pairs. So this is a natural thing to look at.

$$ \matrix { A & \overset { \textstyle f } \rightarrow & B \\ \llap f \downarrow & & \downarrow \rlap { k _ 1 } \\ B & \underset { \textstyle k _ 0 } \rightarrow & C } $$

The cokernel pair is a set $ C $ defined as a quotient of the disjoint union $ B \uplus B $; if $ y $ is an element of $ B $, then I'll write $ y _ 0 $ and $ y _ 1 $ for elements of $ B \uplus B $. Then we have an equivalence relation on $ B \uplus B $ according to which $ y _ 0 $ and $ y _ 1 $ are equivalent iff $ y $ belongs to the image of $ f $, and no other equivalences exist besides the reflexive ones. (More explicitly, $ y _ i $ and $ z _ j $ are equal iff $ y = z $ and $ i = j $ or $ y = z $ and $ y = f ( x ) $ for some $ x $.) Then $ C $ is the quotient of $ B \uplus B $ under this equivalence relation. The cokernel pair also comes equipped with two inclusion/quotient maps from $ B $, while I'll call $ k _ 0 $ and $ k _ 1 $; $ k _ i $ maps $ y $ to the equivalence class of $ y _ i $ in $ C $.

Given an element $ x $ in $ A $, $ k _ 0 ( f ( x ) ) = k _ 1 ( f ( x ) ) $ in $ C $ because $ f ( x ) $ is in the image of $ f $. So since $ f $ is an epimorphism, $ k _ 0 = k _ 1 $. This means that for every element $ y $ of $ B $, $ k _ 0 ( y ) = k _ 1 ( y ) $, so $ y $ is in the image of $ f $.

So $ f $ is surjective.

  • $\begingroup$ Love it. So the central parte here is that for a general, not necessarily epi $f$ and any $b\in B$ we have $k_0(b)=k_1(b)$ iff $b \in im(f)$, right? The next step is then to notice that cokernel pairs of epimorphisms are trivial a.k.a. $k_0=k_1$ for all of $B$. $\endgroup$ Commented Jan 12, 2022 at 14:30
  • $\begingroup$ @GerritBegher : Yes, that's right! $\endgroup$ Commented Jan 12, 2022 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.