0
$\begingroup$

I have read that every quotient map is a surjective homomorphism, but I have troubles with proving it.

Can someone explain me why is that so?

Or even further more, under which conditions this map would be an isomorphism?

(I am trying to prove it in a topological context)

  • 1
    Is "homomorphism" a typo, i.e., did you mean "hom**e**omorphism"? The term "quotient map" has different nuances in algebra and in topology; your use of "isomorphism" suggests algebra, but your tags suggest topology. – 2017-01-14

1 Answers 1

0

You have topological spaces $\;X,\,Y\;$ . A map $\;P:X\to Y\;$ is a quotient map if it is continuous and surjective and it induces the topology in $\;Y\;$ , meaning: $\;U\subset Y\;$ is open iff $\;P^{-1}(U)\subset X\;$ is open

We can say that $\;P\;$ is a quotient map if it is continuous, surjective and either a closed or an open map.

An interesting case is when we have an equivalence relation $\;\sim\;$ on X and we take $\;Y:=X/\sim\;$ the quotient space (f equivalence classes), with the topology induced by the usual (canonical) quotient map $\;P:X\to X/\sim\;,\;\;Px:=[x]\;\;,\;\;[x]:=$ the equivalence clasee of $\;x\;$ .

Thus, what is missing for a general quotient map to be an isomorphism (I suppose you mean here topological isomorphism = homeomorphism) is to be injective and the inverse map to be continuous.

  • 1
    Perhaps I misunderstood but; A quotient map need not be open nor closed. – 2017-01-14
  • 0
    @R Of course not, and now I see I wrote "iff" . Thanks, editing – 2017-01-14