Tip of cone $CA=(A\times I)/(A\times \{0\})$ is contained in the interior of $CA$ Suppose $A$ is subspace of $X$, $p=\{[(a,0)]\}$ is the tip of cone $CA=(A\times I)/(A\times \{0\})$.
In algebraic topology, we can use excision theorem to deduce
$\widetilde {H_n}(X\cup CA)\cong H_n(X\cup CA,CA)\cong H_n(X\cup CA-\{p\}, CA-\{p\})\cong H_n(X,A)$.
This requires the tip $\{p\}$ contained in interior of $CA$, but intuitively, the tip of a cone is on the surface of the cone.

Let $q: A\times I \to CA$ be the quotient map, then tip $\{p\}$ is the image of $A\times \{0\}$ under $q$, but any point in $A\times \{0\}$ belongs to the boundary of $A\times I$.
I know indeed $U=\{[a,t]\mid 0 \leqslant t < \frac{1}{2}\}$ is an open neighborhood of tip $p=\{[(a,0)]\}$, but I still can't understand why  quotient map $q: A\times I \to CA$ can map $A\times \{0\}$ which belongs to of the boundary of $A\times I$, i.e. $\partial (A\times I)=(\partial A \times I) \cup (A \times \partial I) = (\partial A \times I) \cup (A \times \{0\}) \cup (A \times \{1\})$ to interior of $CA$.
What's wrong here? And what is the boundary of $CA$?
Thanks for your times and effort.
 A: $A\times [0,.5)/ A \times \{0\}$ is open in $CA$ by definition of the quotient topology. Since preimage of $A\times [0,.5)/ A \times \{0\}$ under the quotient map from $X \sqcup CA$ is itself, this then implies that $A\times [0,.5)/ A \times \{0\}$ is open in the glued space. This means the tip of the cone is in the interior. 
A: It doesn't make sense to talk about the "boundary" of $A\times I$: the boundary as a subset of what space?  "Boundary" is something that is defined for a subset of a topological space, and very much depends on the ambient space.  If you're considering $A\times I$ as a subset of itself, its boundary is the empty set.  Recall that if $Y$ is a topological space and $B\subseteq Y$, then the boundary of $B$ is defined as $\overline{B}\setminus \operatorname{int}(B)$.  (There are other notions referred to as "boundary" such as the boundary of a manifold but those are not relevant here.)
Similarly, when we're talking about the "interior" of $CA$ we mean its interior as a subset of the space $X\cup CA$.  So to prove $p$ is in the interior, we want to find an open subset of $X\cup CA$ which contains $p$ and is contained in $CA$.  That's what the set $U$ you mention is.  The boundary of $CA$ in $X\cup CA$ is (the image of) $\partial A\times\{1\}$ where $\partial A$ is the boundary of $A$ in $X$.  This takes a little bit of work to prove rigorously but intuitively should be clear: if you picture $X\cup CA$ as $X$ with a cone attached over $A$, the part of the cone that does not have any neighborhood in $X\cup CA$ contained in the cone is just the boundary of the base of the cone.
