# Is every homeomorphism a quotient map?

Let $X$ and $Y$ be topological spaces. Let $f: X \rightarrow Y$ be a bijections such that $f(U)$ is open in $Y$ if and only if $U$ is open in $X$. Then $f$ is a homeomorphism and an open map. In particular, it is a surjective open map and, therefore, it is a quotient map, right?

Now let $X$ be a topological space and let $Y$ be the set of point in the topological space $Y$ above. Munkre's states that if $f:X \rightarrow Y$ is a surjective map then there is exactly one topology on $Y$ relative to which $f$ is a quotient map and it's the quotient topology defined by: $U$ is open in $Y$ if $f^{-1}(U)$ is open in $X$.

Therefore, I deduce that if $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a homeomorphism, then $f$ is a quotient map and $Y$ has the quotient topology. This seems fishy. Is it correct or incorrect?

We like to think of a a quotient map as somehow gluing things together, and a homeomorphism corresponds to the trivial case where we glue together nothing. Moreover, in a homeomorphism, we usually think of $Y$ as being a space in its own right, whereas the quotient topology is more like a structure that the function imposes on the space. (Anthropomorphizing, we might say that the function "wants" to be continuous and so it "demands" that $Y$ have a particular topology).
• So, basically, in practice, a quotient space is always obtained by constructing it from a space $X$ by going modulo an equivalence relation and the fact that a homeomorphism is a surjective open map, and therefore a quotient map, is of no use because it's already a homeomorphism. Right? – Sam Mar 6 '15 at 23:05