# How does the size of a topology affect the continuity of maps?

In the textbook that I'm reading, very often the size of topologies are mentioned, usually in the way the quotient topology is discussed in the picture below. Now I know that the size of a topology in a space corresponds to the number of open sets that it has, but how does that affect the continuity of maps into that space?

Furthermore, in the example it says that "the topology of Y is the largest for which $$\pi$$ is continuous". Isn't the largest topology always the discrete topology? • This "size of a topology" is just a mnemonic, not a precise concept. There is no way to measure the "size" of a topology. Nov 29, 2019 at 10:59
• The point here is the maximality, not the size (cardinality being the obvious size measure). Nov 29, 2019 at 12:44

In the co-domain, enlarging the topology leads to fewer continuous functions. Think of enlarging the topology as adding an open set $$E \subset Y$$, which now makes it harder for any given $$f$$ to be continuous, as you now also must have that $$f^{-1}(E)$$ be an open subset of $$X$$. Thus, the largest topology of $$Y$$ that makes $$\pi$$ continuous need not be the discrete topology, since it may be that there is some subset $$F \subset Y$$ such that $$\pi^{-1}(F)$$ is not open in $$X$$.
If $$\pi: X \to Y$$ then more open sets in $$Y$$ make it more unlikely that $$\pi$$ is continuous. So the largest topology making $$\pi$$ continuous is not the discrete topology in general.
• Would the following be a valid assertion? Let V be an open set in $\mathbb{R}^2 / \sim$ such that $\pi^{-1}(V)$ is a point in $\mathbb{R}^2$, equipped with the usual topology. Then this topology in $\mathbb{R}^2 / \sim$ is "too large", in this sense that we're discussing? Nov 29, 2019 at 10:41