proof that a quotient topology is indeed a topology

Question

So I was proving this elementary exercise of proving that a quotient topology is a topology, where we define the quotient topology via surjection. That is, if $f:X\to Y$ is a surjection, then $\tau_Y = \{t:f^{-1}(t)\in\tau_X\}$. I checked the three conditions of topology, and I did not use in the process that $f$ is a surjection! Is it indeed that case that one does not require the condition? If not, in proving which axiom do I need to invoke that $f$ is surjective?

Answer 1

You are correct that any $f$ lets you define such a topology on $Y$.

We only call this topology a "quotient topology" if $f$ is surjective.

This is the dual of the concept of the "subspace topology:"

If $X$ is a topology, and $i:A\to X$ is injective, the we can define the subspace topology $\tau_A$ as $\tau_A=\{i^{-1}(U)\mid U\in \tau_X\}.$

Again, we can define this topology on $A$ for any function $A\to X,$ but it isn't called the subspace topolohy unless the map is injective.

Answer 2

You're right to note that you don't need surjectivity for $\tau_Y$ to define a topology. Indeed, $Y$ in this topology decomposes as the disjoint union $Y =\ $im$(f)\coprod(Y-$im$(f))$ where im$(f)$ has the quotient topology and $Y-$im$(f)$ has the discrete topology.