Impulse invariance vs. DT representation of a CT system: Where is the inconsistency? Suppose you have a continuous-time (CT) system $h_c(t)$, bandlimited to $B$. Your goal is to represent the system as a discrete-time (DT) system $h[n]$, sampled at $f_s \leq 2 B$. Clearly $h[n]$ won't be a faithful representation of $h_c(t)$ because we do not satisfy the sampling theorem. But I am interested in the exact relationship between $h_c(t)$ and $h[n]$ in this case. There are two ways to think about this:
Representing the CT system as DT one
Based on this question (and answer): Representing a continuous LTI system as a discrete one, $h[n]$ represents a low-pass filtered version of $h_c(t)$:
$$
h[n] = \tilde{h}_c(n T_s) ,\\
\tilde{h}_c(t) = \int h_c(\tau) \operatorname{sinc}\left(t-\tau\right) \operatorname{d}\tau
$$
Impulse Invariance
According to https://en.wikipedia.org/wiki/Impulse_invariance, the sampled $h[n]$ corresponds to an aliased $h_c(t)$:
$$
H(e^{-j\omega}) = \frac{1}{T_s} \sum_{k=-\infty}^{\infty} H_c\left(j\frac{\omega}{T_s} + j\frac{2\pi}{T_s}k \right)
$$
I could take the IDTFT:
$$
h[n] = \sum_{k=-\infty}^{\infty} T_s e^{-j2\pi k n} h_c(n T_s)
$$
which is not the same as the low-pass filtered version above.
There no low-pass filtering.
These two approaches seem to give me two different answers on how to relate $h[n]$ and $h_c(t)$ (when sampling theorem is not obeyed).
What is wrong?
The Impulse Invariant method does not promise to represent the frequency response of the  continuous-time system as a lowpass version.  What the Impulse Invariant method does is frequency-alias the frequency response by sliding it by multiples of the sampling frequency and adding up all of the translated copies.