Consider $X:= \mathbb{PC}^n $ the projective space.
It is well known that the integral homology of $X$ is $H_i(X, \mathbb{Z}) = \mathbb{Z}$ is given by: $0 \leq i \leq 2n$ even, and $H_i(X, \mathbb{Z}) =0$ else.
Let $Y \subset X$ be a closed subspace of (Krull) dimension $m=dim(Y)$.
I often heard that $Y$ induces a so called cycle class $[Y] \in H_{2m}(X, \mathbb{Z})$. Or resp it's "dual" $[Y] \in H^{2n-2m}(X, \mathbb{Z})$ via Poincare.
My question is what is the geometrical intuition behind this ring element $[Y] \in H_{2m}(X, \mathbb{Z})$? What is this "cycle class" of $Y$ concretely?
My ideas:
Naively one might guess that if we compute the homology simplicially then $Y$ is nothing but a isomorphic image of a continuous map $f: \Delta^{2m} \to X$ (especially non degenerated where $\Delta^{2m}$ is a $2m$-simplex) as a free generator of singular complex $C_{\bullet}$ in $2m$-th degree.
And $[Y]$ might be just it's class after taking homology?
Is this exactly the geometrical meaning behind the desired identification or is there another interpretation?
It's instructive to understand the fundamental class by looking at the case of manifolds first. $M$ be a manifold of dimension $n$ and $N \subset M$ be a closed oriented submanifold of dimension $k$. Then the inclusion map $\iota : N \hookrightarrow M$ induces a map in homology $\iota_* : \Bbb Z \cong H_k(N) \to H_k(M)$, and we define $[N] = \iota_*(1)$.
Geometrically, one can understand this as follows. If $N$ has the structure of a smooth manifold, one can triangulate $N$, which is the same as setting up a homeomorphism $K \to N$ from a $k$-simplicial complex $K$. Then composing with the inclusion we get a map $K \to M$. Let restriction of this map to the $k$-simplices of $K$ be denoted by $\sigma_1, \cdots, \sigma_N$ and the restriction to the $(k-1)$-simplices be denoted by $e_1, \cdots, e_M$; note that these are singular $k$-simplices. I claim there is a canonical choice of $\epsilon_1, \cdots, \epsilon_N \in \{+1, -1\}$ such that $\xi = \sum_i \epsilon_i \sigma_i$ is a singular $k$-cycle. How to obtain this choice? Simply expand out $\partial \xi$ as a formal linear combination of $e_1, \cdots, e_M$ and assign $\epsilon_i$ so that $e_p$ and $e_q$ cancel out in this formal expression if and only if the corresponding $p$-th and $q$-th $(k-1)$-simplices in $K$ are pairwise glued. This can always be done coherently as $K$ is a (oriented!) simplicial complex, and therefore gives a singular $k$-cycle $\xi$ (See Hatcher, ch 2.1., pg 108-109. to get an understanding of these ideas). The fundamental class $[N]$ is the homology class represented by $\xi$ in $H_k(M)$.
If $X \subset \Bbb{CP}^n$ is a complex projective variety this is slightly harder. It's a theorem of Whitney that such objects always admit something called a "Whitney stratification", and combining with the result of Mather that Whitney stratified spaces admit something called an "abstract stratification" and that again with Goresky's saying abstractly stratified spaces admit triangulation (see this answer and the references therein), we get a triangulation $K \to X$ for $X$. Now we can play the same game as above to get a singular $2m$-cycle corresponding to $X$, which is the required fundamental class $[X] \in H_{2m}(\Bbb{CP}^n)$