1
$\begingroup$

I want to show that equalisers are unique up to unique homeomorphism. In other words, I want ot see that if $f_1: Y_1 \to X$ and $f_2: Y_2 \to X$ are both equalisers of $g_1, g_2: X \to Z$, then there exists a unique homeomorphism $h: Y_1 \to Y_2$ with $f_1 = f_2 \circ h$.

Any help would be appreciated.

$\endgroup$
6
  • $\begingroup$ Are you familiar with the universal property of the equaliser? $\endgroup$ Feb 22, 2016 at 22:19
  • $\begingroup$ I think no. @StefanHamcke $\endgroup$
    – user189013
    Feb 22, 2016 at 22:21
  • $\begingroup$ This has little to do with general topology, but with catgeory theory, see for example 640149. Even if you haven’t defined equalizers in that way, look up the categorical definition to have your eyes opened. Or maybe, if you have the time, start with the categorical definition of initial and terminal objects and then work your way up to products, coproducts, equalizers … It’s worth it (or even necessary) in the long run. $\endgroup$
    – k.stm
    Feb 22, 2016 at 22:21
  • $\begingroup$ Hints: write down carefully what the definition of an equaliser is telling you about $f_1$ and $f_2$. You will find that the assumption that $f_1$ is an equaliser provides data that you can feed into the the assumption that $f_2$ is an equaliser (and vice versa with $f_1$ and $f_2$ swapped). The uniqueness property in the definition of equaliser will then give you that certain morphisms are identities. This is a standard pattern of for proving that objects defined by universal properties are unique up to isomorphism, e.g, for showing that two free groups on $n$ generators are isomorphic. $\endgroup$
    – Rob Arthan
    Feb 22, 2016 at 22:22
  • $\begingroup$ Given any map $h:W\to X$ such that $g_1h=g_2h$, there is a unique map $k:W\to Y_1$ such that $f_1k=h.$ You have either defined an equaliser this way, or it is an easy consequence of your definition. By the way, it would help to include the definition you are using into your post. $\endgroup$ Feb 22, 2016 at 22:33

1 Answer 1

2
$\begingroup$

This is a typical use of the universal property of the equaliser: if $E$ and $E'$ are both equalisers of $f,g:X \to Y$, with respective morphisms $\mathit{eq}:E \to X$ (resp. $\mathit{eq}':E' \to X$), then (somewhat following the Wikipedia notation) there are unique morphisms $u:E' \to E$ and $u':E \to E'$ such that $\mathit{eq} \circ u = \mathit{eq}'$ and $\mathit{eq'} \circ u' = \mathit{eq}$. In fact these are mutually inverse isomorphisms: we have $\mathit{eq} \circ (u \circ u') = \mathit{eq}' \circ u' = \mathit{eq}$ and $\mathit{eq}' \circ (u' \circ u) = \mathit{eq} \circ u = \mathit{eq}'$, and so by uniqueness we conclude that $u \circ u' = \mathit{id}_E$ and $u' \circ u = \mathit{id}_{E'}$.

The same argument shows more generally that objects defined by a universal property are unique up to a unique isomorphism in any category.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .