Def. for a continuous function:

Let $X$ and $Y$ be topological spaces. A function $f : X \rightarrow Y$ is continuous if $f^{-1} (V)$ is open in $X$ for every open set $V \subseteq Y$.

Def. for a quotient map:

Let $X$ be a topological space and $A$ be a set (that is not necessarily a subset of $X$). Let $p : X \rightarrow A$ be a surjective map. Define a subset $U$ of $A$ to be open in $A$ if and only if $p^{-1} (U)$ is open in $X$. The resultant collection of open sets in $A$ is called the quotient topology induced by $p$ , and the function $p$ is called a quotient map.

In addition to surjectivity of a quotient map, are there any differences between this two concepts? From many different examples in comparison, they seem to be different 'topics' but actually no other difference more than surjectivity in case of a quotient map (?)

Also the book claims that a quotient map is always a continuous function.

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    $\begingroup$ A surjective continuous map $X\to A$ need not be a quotient map, if that's what you're asking. $\endgroup$ – Stefan Hamcke Jun 16 '15 at 12:51
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    $\begingroup$ In general, for a surjective continuous map $X\to Y$, there are subsets $M\subset Y$ such that $f^{-1}(M)$ is open although $M$ is not open. $\endgroup$ – Daniel Fischer Jun 16 '15 at 12:52
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    $\begingroup$ A quotient map is always a surjective continuous function, by definition. But the class of continuous surjections is much broader than that of quotient maps. $\endgroup$ – Stefan Hamcke Jun 16 '15 at 12:55
  • $\begingroup$ @Stefan Hamcke: in latter comment "a quotient map is always a surjective continuous function" but in the first "a surjective continuous function may not be a quotient map". How does it work? $\endgroup$ – L.G. Jun 16 '15 at 12:59
  • $\begingroup$ I got it! Thank you both. $\endgroup$ – L.G. Jun 16 '15 at 13:08

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