Will the width of water leaving the outlet section gradually decrease? 
This is a schematic diagram of a nozzle. Its upper and lower surfaces are parallel, its vertical two sides are not parallel, and the width of the nozzle gradually decreases. Blue represents water ejected from the nozzle.

This is the top view of the nozzle. Blue represents the water ejected from the nozzle.
My question is: will the width of water leaving the outlet section gradually decrease? Will the pressure of water leaving the outlet section be higher than atmospheric pressure? If we ignore the influence of gravity.
I think the answer is yes. Because it's like two streams of water colliding together. Two streams of water bounded by the axis of symmetry of the nozzle.
People ignore inertia. It takes force for water to change the direction of velocity. Why does the water from the nozzle outlet section change its velocity direction?

@ChetMiller
According to @ChetMiller's analysis, I guess the distribution of axial pressure is shown in the red curve in the figure.
Reference:
https://en.m.wikipedia.org/wiki/Vena_contracta
 A: The flow of most fluids through a hose, pipe, etc is not laminar (meaning you cannot assume all the water molecules are traveling in a straight line nearly parallel with  the edge of the nozzle).  Therefore, it is somewhat simplistic to think that the water flow will continue in a straight line path after it is emitted from the nozzle (neglecting gravity as you have suggested) causing the width of the water stream to continue to decrease after it has left the nozzle.
Further, the reason for tapering a nozzle is to increase the flow speed.  This is because an equal volume of water must pass through each section of the hose/nozzle over a fixed interval of time.  Therefore, as you decrease the width of the hose/nozzle you increase the velocity of the water.  Since there is no mechanism to increase the flow rate after the water has left the nozzle, it is unlikely that the width of the band of water will continue to get smaller after being ejected from the nozzle head.
Yes, the total pressure at the nozzle outlet will be greater than atmospheric pressure.  If it were not, the water would not be exiting the nozzle.  Note, that the total pressure is made up of two components:  a static pressure plus a dynamic pressure.  Generally you can think of the static pressure as being equal to the atmospheric pressure and the dynamic pressure as the additional pressure driving the flow of the water.
A: We're going to be looking at a nozzle of circular cross section, that tapers in the axial direction such that its area is still decreasing at the nozzle exit and, as a result, a Vena Contracta forms by the exit jet.  Let the subscript U refer to the region upstream of the nozzle, E refer to the exit cross section, and $\infty$ refer to the final downstream region where the jet area becomes constant again.
Previously we had analyzed this system using a control volume that ran from upstream of the nozzle to the nozzle exit, but we assumed that, at the exit, the pressure is uniform at atmospheric pressure (gauge pressure of zero $p_E=0$) and flow streamlines that are parallel to the nozzle centerline, with a uniform exit velocity $V_E$.  For this approximation, we found that the axial force exerted by the nozzle on the fluid is in the opposite direction of the flow, and given by
$$F^*=\frac{\dot{M}^2A_U}{2\rho}\left[\frac{1}{A_E}-\frac{1}{A_U}\right]^2$$where $\dot{M}$ is the mass flow rate and $\rho$ is the fluid density, and where the superscript * on F indicates that it is the result of our approximation.  With this approximation, we also find that the upstream pressure is given by $$P_U^*=\frac{\dot{M}^2}{2\rho}\left[\frac{1}{A_E^2}-\frac{1}{A_U^2}\right]$$
If we re-analyze this problem by extending the control volume all the way to the final downstream area of the jet (beyond the Vena Contracta) where the pressure actually is atmospheric (and uniform), the axial velocity actually is uniform, and the streamlines are actually parallel to the nozzle axis, we find that the total axial force exerted by the nozzle and outside air (surrounding the exit jet) on the flowing fluid is now given by: $$F=\frac{\dot{M}^2A_U}{2\rho}\left[\frac{1}{A_{\infty}}-\frac{1}{A_U}\right]^2$$ And the upstream pressure is given by: $$P_U=\frac{\dot{M}^2}{2\rho}\left[\frac{1}{A_{\infty}^2}-\frac{1}{A_U^2}\right]$$Note that the force exerted by the outside air on the exit jet most be negligible (neglecting air drag), so F must be the corrected force exerted solely by the nozzle.  So the correction to the force as a result of relaxing the original assumptions and including the Vena Contracta effect must be $$\delta F=F-F^*=\frac{\dot{M}^2A_U}{2\rho}\left[\frac{1}{A_{\infty}}-\frac{1}{A_E}\right]\left[\frac{2}{A_U}+\frac{1}{A_{\infty}}+\frac{1}{A_E}\right]$$This equation indicates that, as a result of including convergence of the nozzle right up to the exit cross section, the calculated force of the nozzle on the flowing fluid is higher than when the effect is neglected.
Similarly, the correction to the upstream pressure is given by $$\delta P_U=P_U-P^*_U=\frac{\dot{M}^2}{2\rho}\left[\frac{1}{A_{\infty}^2}-\frac{1}{A_E^2}\right]$$  This equation indicates that, as a result of including convergence of the nozzle right up to the exit cross section, the calculated upstream pressure is higher than when the effect is neglected.
Next, let's focus on what's happening at the exit cross section.  Here, the exit pressure varies with radial position $P_E=P_E(r)$, the axial component of velocity varies with radial position $V_{zE}=V_{zE}(r)$, and there is a radial component of velocity that also varies with radial position $V_{rE}=V_{rE}(r)$. The average pressure and axial velocity (averaged over the exit cross section) are given by $$\bar{P}_E=\frac{\int_0^{R_E}{2\pi rP_E(r)dr}}{\pi R_E^2}$$and$$\bar{V}_{zE}=\frac{\int_0^{R_E}{2\pi r}V_{zE}(r)dr}{\pi R_E^2}=\frac{\dot{M}}{\rho A_E}$$where $R_E$ is the radius of the nozzle at the exit cross section.  In addition, we express the axial velocity at the exit as the average axial velocity plus the radially varying deviation of the axial velocity from the from the average:  $$V_{zE}=\frac{\dot{M}}{\rho A_E}+\delta V_{zE}(r)\tag{1}$$
Next, we perform a macroscopic momentum balance on the portion of the fluid jet within a control volume between the exit cross section and an arbitrary axial location beyond the Vena Contracta:  $$\bar{P_E}A_E=\rho V_{\infty}^2A_{\infty}-\rho \int_0^{R_E}{2\pi rV_{zE}^2(r)dr}$$If we substitute Eqn. 1 into this equation, we obtain:$$\bar{P_E}A_E=\rho V_{\infty}^2A_{\infty}-\frac{\dot{M}^2}{\rho A_E}-\rho \overline{(\delta V_{zE})^2}A_E\tag{2}$$where $$\overline{(\delta V_{zE})^2}=\frac{\int_0^{R_E}{2\pi r(\delta V_{zE}(r))^2dr}}{\pi R_E^2}$$The first term on the right hand side of Eqn.2 can be re-expressed as $$\rho V_{\infty}^2A_{\infty}=\frac{\dot{M}^2}{\rho A_{\infty}}$$Substituting this into Eqn.2 then yields for the average pressure at the nozzle exit $$\bar{P_E}=\frac{\dot{M}^2}{\rho A_E}\left[\frac{1}{A_{\infty}}-\frac{1}{A_E}\right]-\rho \overline{(\delta V_{zE})^2}$$The first term on the right hand side is clearly positive, but the second term is negative, and it is not clear which of these two terms will prevail.  Thus, according to this analysis, it is not obvious whether the average gauge pressure at the nozzle exit will be positive or negative.
A: 
People ignore inertia. It takes force for water to change the direction of velocity. Why does the water from the nozzle outlet section change its velocity direction?

Very soon after leaving the nozzle the stream of water will break into drops. The more convergent or divergent the nozzle the smaller the drops. The force that drives the formation of drops is the surface tension.
