Given a quasi-ordered set $(Q,\leq)$ the *interval topology* on $Q$ is generated by
$$\{Q\setminus\downarrow x : x\in Q\} \cup \{Q\setminus\uparrow x : x\in Q\},$$
where $\downarrow x = \{y\in Q: y\leq x\}$ and $\uparrow x = \{y\in Q: y\geq x\}$.


Let $\mathbb{N}^\mathbb{N}$ denote the set of all functions $f:\mathbb{N}\to\mathbb{N}$ and define a quasi-ordering relation $\leq^\*$ on $\mathbb{N}^\mathbb{N}$ by $$f\leq^\* g \text{ iff } \exists k\in \mathbb{N} \text{ such that } f(n) \leq g(n) \text{ for all } n\geq k.$$


Endow $\mathbb{N}$ with the discrete topology and let $\tau$ be the product topology on $\mathbb{N}^\mathbb{N}$. Let $\tau\_i(\mathbb{N}^\mathbb{N})$ be the interval topology of $(\mathbb{N}^\mathbb{N}, \leq^\*)$.


The identity map $\text{id}: (\mathbb{N}^\mathbb{N},\tau)\to (\mathbb{N}^\mathbb{N},\tau\_i(\mathbb{N}^\mathbb{N}))$ is continuous, which gives rise to a natural question:


Is there a continuous surjection $f: (\mathbb{N}^\mathbb{N},\tau\_i(\mathbb{N}^\mathbb{N}))\to (\mathbb{N}^\mathbb{N},\tau)$?



Interval topology on $(\mathbb{N}^\mathbb{N},\leq^*)$
In view of the answer to [this question](https://mathoverflow.net/questions/226778/is-the-interval-topology-of-mathbbn-mathbbn-leq-connected/226783#226783), the answer is no, for $\tau\_i$ is connected, but $\tau$ is not.


