Proving a set has quotient topology.

Question

Let $f:X\longrightarrow{Y}$, $s:Y\longrightarrow{X}$ be continuous functions where $fs=id$ how can I prove that Y has the quotient topology for $f$?

Answer 1

If you don't want to use the universal property of the quotient, you can argue directly:

$f$ is surjective because $f\circ s=1_Y:$ let $y\in Y.$ Then, $f(s(y))=y$ so $f$ maps $s(y)$ yo $y$.

Now, if $f^{-1}(V)$ is open in $X$, then $s^{-1}(f^{-1}(V))$ is open in $Y$ since $s$ is continuous.

And then, since $f\circ s=1_Y$ we have

$V=f(s(s^{-1}(f^{-1}(V)))=1_Y(s^{-1}(f^{-1}(V)))=s^{-1}(f^{-1}(V))$ so $V$ itself is open in $Y$.

And so the result follows by definition of the quotient topology.

Answer 2

You can use the universal property. The function $s:Y\to X$ is a quotient if and only if it satisfies the following two properties: 1) $s$ is suryective; 2) for every topological space $Z$ and every function $h:X\to Z$ we have that $h$ is continuous if and only if $h\circ s$ is continuous.