BJT output impedance seen from emitter vs collector
I think I see where your question comes from. And a good answer may require a chapter or two and hours of dialog, I think. But I can try and provide a few clues, perhaps.

In the context I think you are asking, the collector is often said to be approximately "a current source" (in a PNP case; or "current sink" for an NPN.) Current sources/sinks are ideally infinite impedance. (Of course, a BJT is not ideal.) But it's also true that the PNP collector can "pull up hard" (or an NPN collector can "pull down hard") and in that sense it seems as though a collector has very low impedance. Both views are true.

If you are thinking about a specific situation where a certain, known base current is being supplied (or \$V_{BE}\$ is known) and if the BJT is in its active region of behavior (\$\beta\$ is relatively constant over the range of collector currents we might consider), then a given base current (or a given \$V_{BE}\$, your call) will imply a specific collector current. This collector current (ignoring the Early Effect due to basewidth modulation) won't change if you try and apply different \$V_{CE}\$ voltages. If you apply \$V_{CE}=4\:\textrm{V}\$ on the NPN BJT, the collector current will be about the same as when you apply \$V_{CE}=10\:\textrm{V}\$. It just won't matter much. This is what's meant when saying that the collector impedance appears to be high. The collector current doesn't appear to be affected by the \$V_{CE}\$ voltage.

If you hook up the collector to a collector resistor (the other side of which goes to a power supply rail), then the output impedance won't be impacted much by the collector itself and will instead be almost entirely determined by the resistor's value. The collector appears to have a very high impedance, once again.

On the other hand, if you saturate the BJT (swamping the base with current) and use it as a switch then the collector will tie itself very close to the emitter and the voltage between them will remain relatively fixed and small in magnitude. In this case, \$V_{CE}\$ looks a lot like a very low-valued battery (say, about \$200\:\textrm{mV}\$.) Well, you know that batteries are ideally "very low impedance." And guess what? In this case, the collector now looks more like a low-impedance battery and the collector has "low impedance." This is why a BJT can also work as a switch.

So context matters. In active mode, with a known \$V_{BE}\$ (or known base current) the \$V_{CE}\$ terminal pair "looks" like a current source and has high impedance. In deep saturation mode, the \$V_{CE}\$ terminal pair "looks" like a tiny voltage source and has low impedance.



In a forward-biased BJT, the emitter is almost always just a diode drop away from the base. So the emitter follows the base voltage. If the base voltage is tied to a relatively "stiff" source (stiff, relative to the required base current) and if the collector is tied to a good source of current (like a power supply rail in common collector mode), then a load on the emitter (something attached between the emitter and some voltage rail) will see what appears to be a very fixed voltage at the emitter. Even if the load requires a lot more current, the change in the emitter voltage will only be about \$60\:\textrm{mV}\$ for a 10-fold increase in load current at the emitter. So the emitter will look, so far as the load is concerned, very stable and very much like a voltage source. Which means "low impedance."