I'm having a mind-blank right now, and I can't seem to find an explanation of this in the texts I've consulted. It's to do with the uniqueness up to isomorphism of equalizers.

The proof goes:

Let $e: E \rightarrow A$ and $e^\prime : E^\prime \rightarrow A$ be equalizers of morphisms $f$ and $g$ from $A$ to $B$. Then there is a unique $k : E^\prime \rightarrow E$ and a unique $h:E\rightarrow E^\prime$ such that $$e^\prime = e \circ k$$ and $$ e = e^\prime \circ h,$$ thus $$e^\prime = e \circ k = e^\prime \circ h\circ k$$ and $$ e = e^\prime\circ h = e \circ k \circ h. $$

My question is:

What then do we invoke to say that this implies $h\circ k$ and $k\circ h$ are the identity morphisms?

Adamek-Herrlick-Strecker in their Abstract and Concrete Categories (from which this is taken) merely say "by the uniqueness requirement" in the definition of equalizers, which I do not see how to apply.

  • 2
    $\begingroup$ Equalizers are final objects of a certain category built from the original. Thus they are unique up to isomorphism. $\endgroup$
    – dfeuer
    Jan 16, 2014 at 7:13
  • $\begingroup$ Sure that would of course give the result, but I'd really like to know about this particular proof, i.e. what lets us say $h\circ k$ and $k\circ h$ are the identities? $\endgroup$
    – Josh
    Jan 16, 2014 at 7:13
  • 1
    $\begingroup$ The result you need is that equalizers are monic which is proven here: math.stackexchange.com/questions/81296/every-equalizer-is-monic. $\endgroup$ Jan 16, 2014 at 7:22
  • $\begingroup$ I was just looking at that actually... working on it, thanks! $\endgroup$
    – Josh
    Jan 16, 2014 at 7:23

1 Answer 1


Consider the diagram formed from the morphisms $e : E \rightarrow A$ and another copy of $e: E \rightarrow A$. By definition, there is a unique morphism $E \rightarrow E$ which makes this diagram commute. But you proved that $k \circ h$ and the identity are two such morphisms, so they are equal. Same for $E'$.


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