The setting is the one of algebraic curves over the complex numbers. It is known that in an irreducible nodal curve each node reduces the arithmetic genus by one: if $\tilde{C} \rightarrow C$ is the normalization of $C$, and $C$ is nodal with $n$ nodes, then $p_a(\tilde{C})=p_a(C)-n$. I am using the word reduce because I am mainly thinking of a plane curve: if it has degree $d$ and it is smooth, then it has genus \begin{equation} p_g=p_a=\frac{(d-1)(d-2)}{2}, \end{equation} while if it is nodal with $n$ nodes, the genus of the normalization is \begin{equation} p_g=p_a=\frac{(d-1)(d-2)}{2}-n. \end{equation}
The above statement can be proved thinking of our irreducible nodal curve $C$ as being in some smooth surface $S$, using adjunction formula on $S$ and on its blow up at the nodes of $C$.
My question is the following. Can we show this fact using the topological Euler characteristic? Does the topological Euler characteristic make sense if $C$ is singular, and is it still equal to $2(1-p_a(C))$?
Edit: PVAL's comment shows that the above thing does not work for reducible curves. If we take as definition of $p_a$ the one given in Hartshorne (chapter 3, section 5, exercises), then $p_a(C)=h^1(\mathcal{O}_C)$ for a reducible curve too. If we consider $C=L_1 \cup L_2 \subset \mathbb{P}^2$, then we have $h^1(\mathcal{O}_{L_1\cup L_2})=0$ from the long exact sequence in cohomology of $0 \rightarrow \mathcal{O}_{\mathbb{P}^2}(-2)\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow \mathcal{O}_{L_1 \cup L_2}\rightarrow 0$. On the other hand, $\chi_{top}(L_1 \cup L_2)=3 \neq 2 (1-0)$.
The topological Euler characteristic of a singular curve is not always even. In fact, if $\tilde{C}$ is the normalization of $C$, then $\chi(C)=\chi(\tilde C)-n$ where $n$ is the number of nodal points. Indeed, let $x$ be a nodal point of $C$ and $C_x$ be the blow-up of $C$ at $x$. Then, the cartesian and cocartesian square $$\require{AMScd} \begin{CD} \{x_1,x_2\}@>>> C_x\\ @VVV@VVV\\ x@>>>C \end{CD} $$ gives a long exact sequence in cohomology : $$ \dots\rightarrow H^n(C)\rightarrow H^n(C_x)\oplus H^n(x)\rightarrow H^n(\{x_1,x_2\})\rightarrow H^{n+1}(C)\rightarrow\dots $$ In particular $\chi(C)+\chi(\{x_1,x_2\})=\chi(\{x\})+\chi(C_x)$. And because the characteristic of a point is $1$, $\chi(C)=\chi(C_x)-1$. By induction on the number of nodes, you get $\chi(C)=\chi(\tilde{C})-n$.