Is a geodesic in the 4d spacetime still a geodesic after projection onto the 3d space? According to Einstein's general relativity, a free object traces a geodesic in the 4d curved spacetime. But what we can observe is the projected path of the object onto the 3d space, which is itself a Riemannian manifold, with the inherited Riemannian metric from the spacetime. 

Is the projected geodesic onto the 3d space still a geodesic?

We know in a flat space, a straight line projected to a subspace remains straight. But I'm not sure whether this is true for a geodesic in a curved space. I guess my question is how the geodesic equation in the spacetime
$$\frac{d^2x^{\mu}}{ds^2} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{ds}\frac{dx^{\beta}}{ds}=0$$
changes when we only look at components $x^1,x^2,x^3$ (with $x^0 $ being the time $t$) in the 3d space.
Edit 1:
I forgot to mention my background. I am familiar with Riemannian geometry, but I haven't studied general relativity yet besides some casual reading here and there. But this question was stuck in my head long enough so I guess this is the time I let the demon come out. I will appreciate if you can explain your answer in a level suitable for a beginner in general relativity.
 A: The spatial metric associated to some familly of local observers is the following (in some coordinates system.  I'm using signature $\eta = (1, -1, -1, -1)$):
$$\tag{1}
h_{\mu \nu} = u_{\mu} \, u_{\nu} - g_{\mu \nu},
$$
where $u^{\mu}$ are the components of the 4-velocity of the local observer.  Components (1) define a projector:
$$\tag{2}
h_{\mu \lambda} \, h^{\lambda}_{\; \nu} = h_{\mu \nu}.
$$
Also: $h_{\mu \nu} \, u^{\nu} \equiv 0$ and obviously $h_{\mu}^{\; \lambda} \, g_{\lambda \kappa} \, h^{\kappa}_{\; \nu} \equiv h_{\mu \nu}$.  The 3D spatial section defined with this lower metric depends on the familly of observers you select.
You could then compute the 3D Riemann tensor on that lower space.  The spacelike geodesics of that lower space have nothing to do with the timelike geodesics of the full 4D metric, despite that (1) is a projector.
Notice that the lower metric (1) isn't compatible with the full connection:
$$\tag{3}
\nabla_{\mu} \, h_{\lambda \kappa} = \nabla_{\mu} (u_{\lambda} \, u_{\kappa}) \ne 0.
$$
To define your lower Riemann tensor and also the 3D geodesics, you'll need a new connection $\tilde{\Gamma}_{\mu \nu}^{\lambda}$ from $h_{\mu \nu}$ such that
$$\tag{4}
\tilde{\nabla}_{\mu} \, h_{\lambda \kappa} = 0.
$$
A: 
But what we can observe is the projected path of the object onto the 3d space, which is itself a Riemannian manifold, with the inherited Riemannian metric from the spacetime.

Not true. To get a 3d space, you have to pick a surface of simultaneity. GR doesn't in general have a preferred surface of simultaneity.

Is the projected geodesic onto the 3d space still a geodesic?

No. For example, the earth orbits the sun, and the sun's field is approximately static, so that there is a preferred time-slicing. With this time-slicing, space is nearly flat, and the earth's orbit is an ellipse, which is not a geodesic.
