Any line in $\mathbb{P^2}$ is isomorphic to $\mathbb{P^1}$. Any line in $\mathbb{P}^2$ is isomorphic to $\mathbb{P}^1$. I have a doubt we have to show isomorphism topologically or vector space or anything else?
I have found a map which is the following:
Let $X=\{[x:y:z]:F(x,y,z)=ax+by+cz=0\}$  ,where $a,b,c$ not all zero, denote a line in $\mathbb{P}^2$ . Now define a map from $X$ to $\mathbb{P}^1$  assuming $a\neq 0$ by sending $[m:n]$ to $[-(bm+cn)/a:m:n]$.
To show this map is an isomorphism  I have to show this h0lomorphic and and has a holomorphic inverse, basically I have to show isomorphic as Riemann surfaces.  If we take the inverse map $[x:y:z]$ goes to $[y:z]$ then this will give an isomorphism.
Am I correct ?
 A: Otto Forster in his Lectures on Riemann Surfaces calls two Riemann surfaces isomorphic if there exists a biholomorphism between them. Said differently, two Riemann surfaces are (holomorphically) isomorphic if they are isomorphic in the category of Riemann surfaces together with holomorphic maps.
In Forster's terminology, let me show you that any line in $\mathbb{P}^2$ is isomorphic to $\mathbb{P}^1$. With holomorphicity suitably defined, a holomorphic map between Riemann surfaces is continuous. In particular, any line in $\mathbb{P}^2$ is homeomorphic (what you call "topologically isomorphic") to $\mathbb{P}^1$.

Assume without loss of generality that $a\neq 0$. The cases $b\neq 0$ and $c \neq 0$ are treated analogously.
Consider the projection $$\pi\colon X \longrightarrow \mathbb{P}^1 \\ [x:y:z]\longmapsto [y:z].$$ Since $a\neq 0$  holds, we have $[1:0:0]\notin X$. Thus, the mapping $\pi$ is indeed defined on the whole of $X$.
Next consider, as you did, the mapping
$$\iota \colon \mathbb{P}^1 \longrightarrow X \\ [y:z]\longmapsto [-\frac{by+cz}{a}:y:z].$$
Verify that $\iota$ is the two-sided inverse of $\pi$. This makes $\pi$ a bijection.
A bijective holomorphic mapping between Riemann surfaces is biholomorphic by the Local Normal Form of non-constant holomorphic mappings between Riemann surfaces. Thus, it remains to show that $\iota$ (or equivalently $\pi$) is holomorphic.
How to do so depends on your definition of the complex structure on $X$.

*

*If you define the complex structure simply as the one induced from the complex projective line $\mathbb{P}^1$ by the bijection $\pi$, there is of course nothing to show.


*Another (a priori different, but ultimately equivalent) definition of a complex structure on $X$, might be phrased along the following lines. Since $a\neq 0$, the homogeneous polynomial $F(x,y,z)=ax+by+cz$ is non-singular. We can thus consider $X$ as a smooth projective plane curve. The complex structure of a general smooth projective plane curve is described in detail on pages 13-16 of Rick Miranda's Algebraic Curves and Riemann Surfaces.
In our case, Miranda's definition amounts to the subsequent complex structure on $X$. Define the two sets $U_1\colon =\{[x:y:z]\in \mathbb{P}^2 \vert \ y\neq 0\}$ and $U_2\colon =\{[x:y:z]\in \mathbb{P}^2 \vert \ z\neq 0\}$. Since $a\neq 0$, we can cover $X$ by the sets $X\cap U_1$ and $X\cap U_2$, which are open in $X$. The charts of $X$ are constructed by employing the Implicit Function Theorem. More explicitly, if $p\in U_1\cap X$, consider the dehomogenization $F(u,1,v)$. Then we have $\frac{\partial F}{\partial u}(u,1,v)=a\neq 0$, and a chart of $X$ near $p$ is given by $\psi_1([x:y:z])=z/y$. Similarly, if $p\in U_2\cap X$, then $\psi_2([x:y:z])=y/z$ defines a chart of $X$ near $p$.
Now, an atlas of $\mathbb{P}^1$ is given by $\phi_1\colon V_1\rightarrow \mathbb{C}; [z:w]\mapsto w/z$ and $\phi_2\colon V_2\rightarrow \mathbb{C}; [z:w]\mapsto z/w$, where $V_1\colon=\{[z:w]\in \mathbb{P}^1\vert z\neq 0\}$ and $V_2\colon=\{[z:w]\in \mathbb{P}^1\vert w\neq 0\}$. The inverses of $\phi_1$ and $\phi_2$ read as $\phi_1^{-1}(w)=[1:w]$ and $\phi_2^{-1}(z)=[z:1]$, respectively. Observe that, wherever it is defined, for $i,j\in \{1,2\}$, the composition $\psi_i\circ \iota \circ \phi^{-1}_j$ is equal either to the holomorphic function $\mathbb{C}\setminus\{0\}\rightarrow \mathbb{C}\setminus\{0\};z\mapsto 1/z$ or to the holomorphic function $\mathbb{C}\rightarrow \mathbb{C};z\mapsto z$. Thus, the mapping $\iota$ is holomorphic.
A: Another less explicit argument involving slightly more advanced tools goes as follows.
The smooth projective plane curve $X$ has degree $1$.
By Plücker's formula, it thus has genus $0$. Any smooth projective plane curve is a compact Riemann surface. Since any compact Riemann surface of genus $0$ is isomorphic to the complex projective line (I will give a precise reference later), it follows that $X$ is isomorphic to the complex projective line.
As a side remark, the argument above is kind of circular. At least, the proof of Plücker's formula which I know uses the fact that for any smooth projective plane curve $Y$ of degree $d$ one of the three projections $Y\rightarrow \mathbb{P}^1$, e.g. $[x:y:z]\mapsto [y:z]$, is a well-defined holomorphic mapping of degree $d$. In our case, this is precisely the claim that there exists a holomorphic mapping $X\rightarrow \mathbb{P}^1$ of degree $1$, i.e. an isomorphism.
