JupyterLab - sharing kernels breaks ipympl

Dieses Issue hat noch niemand übernommen.

## Describe the issue
If I have two notebooks open in jupyterlab, with the kernel of the second chosen as the kernel of the first (using jupyterlab gui kernel dropdown menu -> change kernel and making a choice from "use kernel from preferred session" and selecting notebook1.ipynb). If I try and use ipympl in this second notebook I get the "Error displaying widget: model not found" error. But ipympl works perfectly if I select the second (and original) kernel for the second notebook.

I'm doing this kernel sharing because I am using jupyterlab to control my experiment and only once instance of the hardware control code can be loaded. My idea is that my various measurements will be in separate jupyter notebooks but control a single kernel (where the hardware is defined). So far this idea works fine except for ipympl, I can still plot with `%matplotlib inline`.

I have tested this in a fresh conda environment with the latest versions of everything (AFAIK).

## Versions

```
3.9.5 (default, May 18 2021, 14:42:02) [MSC v.1916 64 bit (AMD64)]

ipympl version: 0.7.0
jupyter core : 4.7.1
jupyter-notebook : 6.4.0
qtconsole : 5.0.3
ipython : 7.22.0
ipykernel : 5.3.4
jupyter client : 6.1.12
jupyter lab : 3.0.16
nbconvert : 6.0.7
ipywidgets : 7.6.3
nbformat : 5.1.3
traitlets : 5.0.5
Known nbextensions:
config dir: C:\Users\monnvadmin\anaconda3\envs\control\etc\jupyter\nbconfig
notebook section
jupyter-matplotlib/extension enabled
- Validating: ok
jupyter-js-widgets/extension enabled
- Validating: ok
JupyterLab v3.0.16
C:\Users\monnvadmin\anaconda3\envs\control\share\jupyter\labextensions
jupyter-matplotlib v0.9.0 enabled ok
@jupyter-widgets/jupyterlab-manager v3.0.0 enabled ok (python, jupyterlab_widgets)

```

Reproduziere das Problem in JupyterLab 3.0.16, indem du den Kernel von notebook1.ipynb über das Kernel-Dropdown-Menü freigibst. Verwende dann ipympl im zweiten Notebook und beobachte den Widget-Modellfehler. Vergleiche dies mit dem funktionierenden Fall mit separaten Kernels; abgeschlossen ist die Aufgabe, wenn ipympl korrekt angezeigt wird, wenn Kernel gemeinsam verwendet werden, ohne den gemeldeten Fehler.

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