Time in Universe Chronology - Duration of inflation period When you read articles about the chronology of the Universe, you read for example that the duration of the inflation era is roughly $10^{-32}$s. We also know   Universe expansion causes  a redshift which can be expressed  in terms of ratio of two frequencies. If we imagine  the comparison of ideal clocks between now and then, knowing the redshift, we could evaluate a time dilation between  proper time at the time of inflation, and time measured on our ideal clock today. My question is the following : when one speaks of a duration of $10^{-32}$s for the inflation, are we talking about   proper time, i.e. the time that an hypothetical ideal clock would measure at that moment, or are we speaking of "Cosmological time", that is  time measured today on our own ideal clock? 
Let me clarify  the reason for my question. If indeed, inflation ends at about t = $10^{-32}$ s, as indicated in this   plot from Wikipedia, and if this time is measured in our present frame where the clocks are, it means that the duration of inflation, as measured by such clock, has an upper bound of $10^{-32}$s. Therefore, an ideal clock located at inflation time, would measure an even shorter time. Assuming as mentionned here that "Cosmic inflation expands space by a factor of the order of $10^{26}$ over a time of the order of $10^{−33}$ to $10^{−32}$ seconds", then adding again another expansion factor of $10^{3}$ between decoupling and now, this would imply that the duration of inflation as measured by a "local" clock  at the beginning of inflation, would be less than $10^{-32}*10^{-29}=10^{-61}$ second! (Here I have used  that $1+z=\frac{\nu_E}{\nu_R}=\frac{dt_R}{dt_E}=\frac{a(t_R)}{a(t_E)}$. See Hobson & al., p.368)
I would have figured out that the inflation duration of $10^{-32}$s would have been the proper time duration, and hence that its duration as measured with our current clocks would have been much longer. Not the opposite!  
Any explanation?
 A: Our clocks cannot measure the duration of inflation directly, since they weren't there at the time. What we can (in principle, in the case of inflation) measure are the signals coming from past events. And yes, if these signals come from near the Big Bang they are massively redshifted and slowed down by the time they get to us.
But the times discussed in cosmology are pretty much always proper times of objects comoving with the expansion (a.k.a. the Hubble flow). So when we say that inflation lasted $10^{-32}\, \mathrm{s}$, that's the time a clock would have measured back then. If we could see it with our own eyes (or an instrument), it would seem to us to have happened much more slowly, thanks to the redshift. In fact, this is exactly what happens with the CMB: radio antennas measure an electric field oscillating around $1.5\times 10^{11}$ times per second. But an observer present when it was emitted would have seen it oscillating a thousand times faster. Same goes for inflation.
Edit: to calculate how much longer inflation looks, a simple approximation would be to consider a time interval much shorter than the duration of inflation. In that case you just multiply by the redshift factor, which is indeed around $10^{26}$.
If you want to see what happens to the whole of inflation then things get a bit more complicated, because the redshift factor changes quickly throughout inflation. We use the basic equation $dr = c\, dt/a(t)$ for light rays and consider a point at some comoving distance $R$ from us which emits light at times $t_{e1}$ and $t_{e2}$ (the beginning and end of inglation), which we receive at times $t_{o1}$ and $t_{o2}$. Integrating both sides of the equation from $t_{e1}$ to $t_{o1}$ and from $t_{e2}$ to $t_{o2}$, we find
$$\int_{t_{e1}}^{t_{o1}} \frac{dt}{a} = \int_{t_{e2}}^{t_{o2}} \frac{dt}{a},$$
which implies 
$$\int_{t_{o1}}^{t_{o2}} \frac{dt}{a} = \int_{t_{e1}}^{t_{e2}} \frac{dt}{a}.$$
For the LHS, assume the time difference between the two receptions is shorter than the Hubble time, so we just get $\Delta t_o = t_{o2}-t_{o1}$ (also setting $a(t_0) = 1$). The RHS turns out to be
$$\frac{e^{H \Delta t_e}-1}{a_{BB}H},$$
where $a_{BB}$ is the scale factor at the end of inflation and $H$ is the Hubble parameter during inflation. Plugging in some numbers I found in this answer just as an example, I find $\Delta t_o \approx 5\times 10^{-7}\,\mathrm{s}$.
A: The Big Bang time line is seen here:

The time is counted from the calculated big bang point, but it is given by our present frame where the clocks are. The inflationary epoch is seen in this plot to be between about $10^{-38}$ seconds  to $10^{-36}$ seconds.
In this plot it is seen that a fuzzy region  is drawn for the original  Big Bang estimate  point singularity .
In this article 

The comoving time coordinate is the elapsed time since the Big Bang according to a clock of a comoving observer and is a measure of cosmological time. The comoving spatial coordinates tell where an event occurs while cosmological time tells when an event occurs. Together, they form a complete coordinate system, giving both the location and time of an event.

Comoving time and proper time are two different mathematically numbers, but the timelines of the Big Bang are given in comoving time. This is as far as I have managed to clarify for myself. 
