The current can still flow through the "substrate" even though the channel is pinched. The reason why it saturates is that there will be a region of higher resistance of size proportional to the Drain-Source voltage, and therefore the resistance of this region will be proportional to the same voltage.
But as current is voltage/resistance, the dependence will cancel out and you'll get "constant" current.
From Wiki (emphasis mine):
Even though the conductive channel formed by gate-to-source voltage no longer connects source to drain during saturation mode, carriers are not blocked from flowing. Considering again an n-channel enhancement-mode device, a depletion region exists in the p-type body, surrounding the conductive channel and drain and source regions. The electrons which comprise the channel are free to move out of the channel through the depletion region if attracted to the drain by drain-to-source voltage. The depletion region is free of carriers and has a resistance similar to silicon. Any increase of the drain-to-source voltage will increase the distance from drain to the pinch-off point, increasing the resistance of the depletion region in proportion to the drain-to-source voltage applied. This proportional change causes the drain-to-source current to remain relatively fixed, independent of changes to the drain-to-source voltage, quite unlike its ohmic behavior in the linear mode of operation. Thus, in saturation mode, the FET behaves as a constant-current source rather than as a resistor, and can effectively be used as a voltage amplifier. In this case, the gate-to-source voltage determines the level of constant current through the channel.
Also, from the MOSFET operation description, under saturation:
Since the drain voltage is higher than the source voltage, the electrons spread out, and conduction is not through a narrow channel but through a broader, two- or three-dimensional current distribution extending away from the interface and deeper in the substrate. The onset of this region is also known as pinch-off to indicate the lack of channel region near the drain. Although the channel does not extend the full length of the device, the electric field between the drain and the channel is very high, and conduction continues.
What do you mean by "input characteristics"?
Textbooks and datasheets describe the behavior of MOSFETs using two graphs:
Output characteristics: \$I_D\$ versus \$V_{DS}\$ with \$V_{GS}\$ as parameter.
Transfer characteristic: \$I_{D}\$ versus \$V_{GS}\$ at a given fixed \$V_{DS}\$ value (this latter is chosen so that the MOSFET is in saturation region).
There is no "input characteristic" (such as the \$I_B\$ versus \$V_{BE}\$ curve of a BJT) because the other input quantity besides \$V_{GS}\$, namely \$I_G\$, is virtually zero at DC (and all these curves assume DC operations). Therefore it wouldn't make much sense to plot \$I_G\$ versus \$V_{GS}\$, unless you wanted to analyze leakage gate current, but I assume you are not interested in that.
So it is clear (also by a comment of yours) that by input characteristic you mean the transfer characteristic (TC). Note that the TC is plotted with a fixed drain-source voltage that guarantees that the MOSFET is in saturation for each \$V_{GS}\$ value on the horizontal axis. This is done because the TC is useful when the MOSFET is in saturation, i.e. when the output current depends solely on the input voltage (not considering "Early effect"), for example when you want to use the MOSFET as an amplifier and you need to draw a load line to design its bias circuit.
If you plot the TC for different values of \$V_{DS}\$ you get a family of TC curves. For example consider this circuit simulation with LTspice:
Plotting the TC for different \$V_{DS}\$ values you get:
As you can see, the more you increase \$V_{DS}\$ the more the curve resembles a parabola, as you would expect for the TC in saturation. Notice that this part shows a threshold voltage \$V_{th} \approx 4V\$.
Let's consider what happens if \$V_{DS}\$ is not big enough to drive the MOSFET in saturation for every \$V_{GS}\$ value, like in the lowest blue curve (Note: to present a more revealing plot I selected the curve corresponding to \$V_{DS} = 2V\$, whereas the lowest blue curve above corresponds to \$V_{DS} = 1V\$):
As you can see, in saturation region you get a quadratic curve, whereas in triode region you get a linear curve. Everything as expected, except that real devices don't have an abrupt change between the two regions and that the linearity of the triode region is not perfect because of the device not being ideal (SPICE models usually take into account these effects).
If you see in your simulation an abrupt departure from this behavior it could be that you tried plotting the curves outside the range of the voltages/currents admissible for your device. Notice that I limited the first plot to max 14A/20V which are the absolute maximum ratings for the device I chose. If you don't keep this in mind you will destroy the device (in real life) or get odd results (in simulations).
EDIT (in response to a comment and a question edit)
You ask why the "perfectly" linear curve for \$I_D\$ versus \$V_{GS}\$ in ohmic region is not exploited. Here is some insight:
Why do you need a linear characteristic between input (\$V_{GS}\$) and output (\$I_D\$)? Usually to use the device as a (linear) amplifier. But what are the conditions that allows to have that linearity? \$V_{DS}\$ must be held constant. Therefore to make an amplifier this way you have to insert a load in the output circuit and still keep \$V_{DS}\$ constant. You can understand that such a load cannot be a simple resistor (which is the simplest kind of load). Therefore you need a much more complex circuit (with other active devices).
On the other side, you can use the same MOSFET biased in saturation and get a decent linear amplifier: even if the behavior of the device is not intrinsically linear, but quadratic, there are linearization techniques (e.g. employ simple feedback schemes, like a resistor in series with the source terminal) that allow the overall amplifier to become more linear.
Best Answer
The channel isn't completely cut off due to the strength of the electric field between the drain and the channel. We use this region because the current is no longer dependent on Vds and we can amplify waveforms without distortion as long as we make care to stay in saturation.
I mean strictly speaking you're right the depletion region around the drain in saturation is smaller than it is in the triode region, so it technically isn't as good, but as for the reason stated above it doesn't matter, current is still flowing, but now largely without regard to Vds.