In an exercise in his algebraic topology book, Munkres asserts that $\mathbf{C}P^n$ is triangulable (i.e., there is a simplicial complex $K$ and a homeomorphism $K \rightarrow \mathbf{C}P^n$). Can anyone provide a reference or a proof?
I will present a triangulation of $\mathbb{CP}^{n1}$. More specifically, I will give an explicit regular CW structure on $\mathbb{CP}^{n1}$. As spinorbundle says, the first barycentric subdivision of a regular CW complex is a simplicial complex homeomorphic to the original CW complex. Recall that to put a regular CW complex on space $X$ means to decompose $X$ into disjoint pieces $Y_i$ such that: (1) The closure of each $Y_i$ is a union of $Y$'s. (2) For each $i$, the pair $(\overline{Y_i}. Y_i)$ is homemorphic to $(\mbox{closed}\ d\mbox{ball}, \mbox{interior of that}\ d\mbox{ball})$ for some $d$. The barycentric subdivision of $X$ corresponding to this regular CW complex is the simplicial complex which has a vertex for each $Y_i$ and has a simplex $(i_0, i_1, \ldots, i_r)$ if and only if $\overline{Y_{i_0}} \subset \overline{Y_{i_1}} \subset \cdots \subset \overline{Y_{i_r}}$. Write $(t_1: t_2: \ldots: t_n)$ for the homogeneous coordinates on $\mathbb{CP}^{n1}$. For $I$ a nonempty subset of $\{ 1,2, \ldots, n \}$, let $Z_I$ be the subset of $\mathbb{CP}^{n1}$ where $t_i=t_{i'}$ for $i$ and $i' \in I$ and $t_i > t_j$ for $i \in I$ and $j \not \in I$. Note that $Z_I \cong (S^1)^{I1} \times D^{2(nI)}$, where $D^k$ is the open $k$disc. Also, $\overline{Z_I} = \bigcup_{J \supseteq I} Z_J \cong (S^1)^{I1} \times \overline{D}^{2(nI)}$ where $\overline{D}^k$ is the closed $k$disc. We now cut those torii into discs. For $i$ and $i'$ in $I$, cut $Z_I$ along $t_i=t_{i'}$ and $t_i =  t_{i'}$. So the combinatorial data indexing a face of this subdivision is a cyclic arrangement of the symbols $i$ and $i$, for $i \in I$, with $i$ and $i$ antipodal to each other. For example, let $I=\{ 1,2,3,4,5 \}$ and write $t_k=e^{i \theta_k}$ for $k \in I$. Then one of our faces corresponds to the situation that, cyclically, $$\theta_1 < \theta_2 = \theta_4 + \pi < \theta_3 = \theta_5 < \theta_1+ \pi < \theta_2 + \pi = \theta_4 < \theta_3 + \pi = \theta_5 + \pi < \theta_1.$$ This cell is clearly homeomorphic to $\{ (\alpha, \beta) : 0 < \alpha < \beta < \pi \}$. Similarly, each of these cells is an open ball, and each of their closures is a closed ball. We have put a CW structure on the torus. Cross this subdivision of the torus with the open disc $D^{2(nI)}$. The result, if I am not confused, is a regular $CW$ decomposition of $\mathbb{CP}^{n1}$. 


I think the comments answer the question, but to give you a reference: Milnor, Stasheff: Characteristic Classes, Chapter 6 They prove that every Grasmann manifold $G_n(\mathbb{R}^m)$ is a CWComplex. (The cells are constructed with Schubert symbols). The complex case works in the same fashion. EDIT: Sorry for the sloppiness! Edit 2: $\oplus$: Perhaps the next sloppiness: The CWstructure of $\mathbb{CP}^n$ obtained by Schubert cells isn't regular (the characteristic map is 2to1) but I think there exists a regular CWstructure. But this might be harder to prove than I thought?! 


An online search yielded a reference to Francis Sergeraert's paper, Triangulations of complex projective spaces, available at http://wwwfourier.ujfgrenoble.fr/~sergerar/Papers/ . But, to quote the author: "The Kenzo program is used to automatically produce triangulations of the complex projective spaces $P^nC$ as simplicial sets, more precisely of spaces having the right homotopy type. The homeomorphism question between the obtained objects and the projective spaces is open." 


Here is an article on explicit triangulation on $CP^n$, http://arxiv.org/abs/1405.2568. 


$CP^{n1}$
, one might try to show that there's some triangulation of the next cell such that after attaching this cell we still have a simplicial complex. The attaching map is the quotient map$S^{2n1} \to CP^{n1}$
, whose fibers are copies of$S^1$
. The inverse image of a point under a simplicial map is always discrete, so this attaching map is definitely not a simplicial map, no matter what simplicial structures you use. So I think John's question is not so trivial. – Dan Ramras Apr 7 '10 at 23:18