This is probably a very simple question, but I'm having some trouble with it. Let $X$ be a CW complex. I am trying to show that $X\times[0,1]$, as a CW complex (so not necessarily product topology), deformation retracts onto $X\times\{0\}$.
If we were working with the product topology, the map $F\colon X\times[0,1]\times[0,1]\rightarrow X\times[0,1]$ given by $$(x,s,t)\mapsto (x,s(1-t))$$ should work.
But for a general CW complex the topology on $X\times[0,1]$ may be finer than that of the product topology. I assume that the map I am looking for is the same map, but how do I show that it is continuous?