Embedding a space in its cone

Question

Let $X$ be a topological space, and $C(X)= (X \times [0 ,1])/(X \times {1} )$,

define $f\colon X \to C(X)$ as $f(x)=[x,t]$ for some fixed $t$ s.t $\ 0\leq t <1$.

I have to show this is a continuous homemorphism onto its image. Injectivity is easy, and continuity comes from the quotient topology on cone.

I have problem showing that this is an open map. I start with an open set $O \subset X$ then $f(O) = \lbrace f(x) \vert x \in O \rbrace = \lbrace [x,t] \vert x\in O \rbrace =\lbrace (x,t) \vert x\in O \rbrace= O \times \lbrace t \rbrace$

How do I show this set is open in the cone?

Answer 1

It is not open in the cone, but in the image of your space with the subspace topology. So you have to show that it is the intersection of an open Set of the cone with the image of $X$.

Answer 2

Note that for $O \subseteq X$ open, $O \times [0,1)$ is open in $X \times I$, and consists of all-singleton classes of $C(X)$ when considered as a subset of $C(X)$, which means, by the definition of the quotient topology, that $O \times [0,1)$ is also open (as a set of classes) in $C(X)$.

Now $f[O] = (O \times [0,1)) \cap (X \times \{t\}) = (O \times [0,1)) \cap f[X]$, which is open in $f[X]$ by the definition of the subspace topology on $f[X]$, by the previous paragraph. So $f$ is open between $X$ and $f[X]$.