Let $(X,E)$ be a CW complex and let $x\in X$. Does there exist a CW decomposition of $X$ having $x$ as a 0-cell?
The book I'm reading says that subdividing the cell containing $x$ gives the desired CW decomposition, with no further details. (that's the whole content of the proof!) Of course anyone can see that subdivision is the key point here so I feel that such an explanation is not really a proof but just some advice.
Some chapters earlier, I solved a similar exercise concerning finite simplicial complexes, i.e. polyhedra. I used an argument similar to the barycentric subdivision of a simpmlicial complex there. However, when we deal with a CW complex $X$, it seems that there is no uniform way to handle the subdivision for all cells, because unlike polyhedra, closure of cells need not meet nicely anymore for CW complexes. I have no idea how to subdivide the cells and give them appropriate characteristic maps.
An detailed proof of the qusetion would be very nice. Please enlighten me.
Here's one simple way to do it. The proof is by induction on the dimension of the cell containing $x$.
Let $e$ be the cell containing $x$, let $n$ be the dimension of $e$, and let $\chi : B^n \to X$ be the characteristic map of $e$. The boundary image $\chi(\partial B^n)$ is contained in the $n-1$ dimensional skeleton $X^{(n-1)}$. Choosing any point $y \in \chi(\partial B^n)$, if $y$ is not already a $0$-cell, we may by induction change the CW structure on $X^{(n-1)}$ so that $y$ is a $0$-cell.
Let $\xi = \chi^{-1}(x)$, a point in the interior of $B^n$. Choose $\eta \in \partial B^n$ to be a point such that $\chi(\eta)=y$. Let $(\xi,\eta) \subset B^n$ be the open line segment in $B^n$ with endpoints $\xi,\eta$.
Remove $e$ from the set of $n$-cells of $X$, add $x$ to the set of 0-cells, add $\alpha = \chi(\xi,\eta)$ to the set of $1$-cells, and add $\chi(\text{int}(B) - (\{\xi\} \cup (\xi,\eta)) = e - (\{x\} \cup \alpha)$ to the set of $n$-cells.
It should be evident that $e - (\{x\} \cup \alpha)$ is homeomorphic to the open $n$-ball, and with a little work you should be able to write down a formula for a characteristic map.