Well definedness of the cone functor

Question

I have a question about the cone space $C(X)$ of a topological space $X$ or more precisely about the cone functor $$ \begin{align*} F: &\mathbf{Top} \to \mathbf{Top}\\ &X \mapsto C(X)\\ &f \mapsto Cf, \end{align*} $$ where $C(X) = (X \times I)/(X \times \{1\})$ and $Cf ([x,t]) = [f(x), t]$. I fail to show that $Cf$ is continuous again. My idea was to do a case distinction into (I) $[y, 1] \notin U$ and (II) $[y, 1] \in U$, where in the latter I tried to split the preimage of $U$ into one part containing the presages of the equivalence classes $[y, t] \in U, t \neq 1$ and another part containing the preimage of $[y, 1]$, which is supposed to be $[x, 1]$. However, the latter is not open, so I cannot conclude. Is there an elegant way to express the preimage of $Cf $ such that its continuity is obvious?

Answer 1

$C(f):C(X) \to C(Y)$ fits into a commutative diagram with $q_X : X \times I \to C(X)$ the (defining) quotient map and $q_Y: Y \times I \to C(Y)$ (ditto), and $f \times \textrm{id}_I: X \times I \to Y \times I$ which is continuous whenever $f$ is (standard properties of product).

Namely as follows : $$Cf \circ q_X = q_Y \circ (f \times \textrm{id}_I)$$

and then the universal property of the quotient (applied to $q_X$, see Properties section of this page) implies that $Cf$ is continuous, because the right hand side composition is continuous.