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Let $C([0,T];\mathbb{R}^d)$ denote the space of continuous functions with the usual supremum norm and given a path $x\in C([0,T];\mathbb{R}^d)$ let $x^s$ denote the stopped path $x(t \land s)$ with $t\in [0,T]$. Now, consider the following quotient space $$X :=[0,T] \times C([0,T];\mathbb{R}^d) \ /\sim \ := Y/\sim$$ where $\sim$ is defined as $(t,x) \sim (s,y)$ if and only if $t=s$ and $x^t = y^s$. This quotient space can be equipped with the metric $$d_\infty([(t,x)]_\sim,[(s,y)]_\sim ) := |t-s| + \sup_{u \in [0,T]}|| x^t(u) - y^s(u)||.$$ Now, if $X$ was equipped with the quotient topology, then the quotient map is continuous (by definition I'd say). But I'm endowing $X$ with the topology induced by $d_\infty$. What can be said about the continuity of the quotient map? Is it still continuous? The quotient topology is a final topology, so if the topology induced by $d_\infty$ is comparable to the quotient topology, then the quotient map is continuous. But how to I know if the topologies are comparable even? It's been awhile since I touched Topology, so any insights will be appreciated :) I feel like I'm missing a trivial observation.

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I reckon I just need to show that the pre-image of an open ball in $(Y/\sim, d_\infty)$ is open in $Y$. So an open ball in $(Y/\sim, d_\infty)$ centred at $[(t,x)]_\sim$ is given by

\begin{align*} B_R([(t,x)]_\sim) =& \big\{ [(s,y)]_\sim \in Y/\sim \ : \ d_\infty([(t,x)]_\sim,[(s,y)]_\sim) <R \big\} = \\ =& \left\{[(s,y)]_\sim \in Y/\sim \ :\ |t-s| + \sup_{u \in [0,T]}|| x^t(u) - y^s(u)|| <R \right\}. \end{align*} Now, if $q$ denotes the quotient map, then the pre-image of $B_R([(t,x)]_\sim)$ is given by $$q^{-1} \left( B_R([(t,x)]_\sim) \right) = \left\{ (s,y) \in Y \ : \ |t-s| + \sup_{u \in [0,T]}|| x^t(u) - y^s(u)|| < R \right\},$$ which seems open. But I'd need a rigorous argument for this last statement.

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  • $\begingroup$ Fwiw I think $(X;d_\infty)$ indeed carries the quotient topology $\endgroup$
    – FShrike
    Feb 1 at 15:21
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    $\begingroup$ Firstly, you need a strict inequality, $d_\infty(a,b)<R$, for it to be an open ball. Secondly, you need only argue that $|t-s|$ and $\sup_{u\in[0,T]}||x^t(u)-y^s(u)||$ are continuous functions $X\to \Bbb{R}_+$, then use that their sum is continuous, and then see that the ball is the preimage of $[0,R)$, which is open in $\Bbb{R}_+$. $\endgroup$ Feb 2 at 12:13
  • $\begingroup$ @SomeCallMeTim Indeed, it should be a strict inequality (edited). Upvoted for the rest of the argument :) Thanks a lot. $\endgroup$
    – Oscar
    Feb 2 at 12:19

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