I have heavily edited the post (including the title), based on a comment by @GregoryArone that my map $f$ is not injective.

Unless I am mistaken somewhere, I have stumbled on a smooth map from $\mathrm{P}^2_{\mathbb{C}}$ into $S^5$ which is one-to-one (Edit: my map turns out to be a branched $2$-to-$1$ continuous map onto a codimension $1$ subset of $S^5$). Thus it induces a topological embedding of the former into the latter (Edit: it does not though, because it is not one-to-one). Is it known that such a topological embedding exists by the way? If so, could someone please point me to the relevant literature? (I am not claiming that my map is an immersion).

Edit 1: I did not expect all that interest in this post. Here is a description of the map.

Define the map $g: \mathrm{P}^1_{\mathbb{C}} \times \mathrm{P}^1_{\mathbb{C}} \to \mathrm{P}^2_{\mathbb{C}}$: $$ ([u_0, u_1], [v_0, v_1]) \mapsto [u_0v_0, u_0 v_1 + u_1 v_0, u_1v_1] $$ (the image is the equivalence class, under scaling, of the symmetric tensor product of the two vectors of homogeneous coordinates). Then $g$ is holomorphic and onto. The symmetric group $S_2$ acts on the domain of $g$ by permuting the $u$-point and the $v$-point, so to speak. The fibers of $g$ are actually the $S_2$ orbits in the domain of $g$.

The (extended) Hopf map $h$ is a smooth map from $\mathbb{C}^2$ onto $\mathbb{R}^3$, defined by

$$h(u_0,u_1) = \left( 2 \operatorname{Re}(u_0 \bar{u}_1), 2 \operatorname{Im}(u_0 \bar{u}_1), |u_0|^2 - |u_1|^2 \right).$$

The group $U(1)$ acts on the domain of $h$ by scalar multiplication, and the fibers of $h$ are the $U(1)$-orbits in the domain of $h$.

Then the map $$ h \times h: \mathbb{C}^2 \times \mathbb{C}^2 \to \mathbb{R}^3 \times \mathbb{R}^3$$ followed by the map $\operatorname{Sym}: \mathbb{R}^3 \times \mathbb{R}^3 \to \operatorname{Sym}^2(\mathbb{R}^3)$ which maps $(x,y)$ to $x \odot y$, gives a map $$k: \mathbb{C}^2 \times \mathbb{C}^2 \to \operatorname{Sym}^2(\mathbb{R}^3),$$ which in turn induces a smooth map $$\tilde{k}: \mathrm{P}^1_\mathbb{C} \times \mathrm{P}^1_\mathbb{C} \to S^5,$$ where the latter is the unit sphere in $\operatorname{Sym}^2(\mathbb{R}^3) \simeq \mathbb{R}^6$. Note that in defining $\tilde{k}$, we have made use of the normalization map $$n: \operatorname{Sym}^2(\mathbb{R}^3) \setminus \{ \mathbf{0} \} \to S^5.$$ Then $\tilde{k}$ is smooth and is moreover invariant under the action of $S_2$ on its domain, which are the fibers of $g$. The fibers of $\tilde{k}$ are actually of the form $$(\mathbf{u}, \mathbf{v}), (j\mathbf{u}, j\mathbf{v}), (\mathbf{v}, \mathbf{u}), (j\mathbf{v}, j\mathbf{u})$$ where $\mathbf{u} = [u_0, u_1]$, $\mathbf{v} = [v_0, v_1]$ are points on $\mathrm{P}^1_\mathbb{C}$ and where $$j[u_0, u_1] = [-\bar{u}_1, \bar{u}_0]$$ represents the antipodal map on the $2$-sphere. Thus $h(j\mathbf{u}) = -h(\mathbf{u})$.

Thus it induces a map $$f: \mathrm{P}^2_\mathbb{C} \to S^5$$ as follows. Given a point $p \in \mathrm{P}^2_\mathbb{C}$, let $w \in g^{-1}(p)$ and map $p$ to $\tilde{k}(w)$.

Then $f$ is well defined, continuous map, from $\mathrm{P}^2_\mathbb{C}$ onto a codimension $1$ subset of $S^5$, with fibers of the form $$[z_0, z_1, z_2], [\bar{z}_2, -\bar{z}_1, \bar{z}_0].$$ The branch locus of $f$ consists of homogeneous quadratic polynomials having roots of the form $\mathbf{u}$ and $j\mathbf{u}$ (I am identifying $\mathrm{P}^2_\mathbb{C}$ with the space of homogeneous quadratic polynomials in two complex variables).

Note that $f$ is smooth on the subset of $\mathrm{P}^2_\mathbb{C}$ having distinct roots, where we think of an element of $\mathrm{P}^2_\mathbb{C}$ as a nonzero complex polynomial of degree at most $2$. I am not sure if $f$ is smooth though at points in the discriminant locus.

Edit 2: one should be able to work out what the image of $f$ is. What is the image of my map $f$? This should be known in algebraic geometry.

I suspect that what I am looking at is related to, for instance, the content of the paper by Atiyah and Berndt https://arxiv.org/pdf/math/0206135.pdf. I suspect I am just rediscovering that the complex projective plane modulo complex conjugation is the $4$-sphere, except that instead of complex conjugation, I am using a different real structure.