thermodynamics
According to widipedia:
During step 1 (isothermal heat addition or absorption) the gas is
allowed to expand and it does work on the surroundings. The
temperature of the gas does not change during the process, and thus
the expansion is isothermal. The gas expansion is propelled by
absorption of heat energy Q1.
The gas temperature is the same as the temperature of the reservoir.
But where may heat energy Q1 come from as there is no temperature gradient between gas and reservoir?
In an isothermal process, as you say, the temperature of the system doesn't change. If your question is "why?" the answer is simple: by definition.
If your question is "how?", then there are two possible answers. The first is it doesn't matter, whatever process you come up with, which makes the temperature constant, will be called isothermal.
A more useful answer, however is that typically this applies to systems which are in contact with a large heat reservoir. So, the "condition" that you mention is given by the zeroth law of thermodynamics, which imposes thermal equilibrium of systems in contact with each other. To ensure this happens during the whole process, the process has to be slow enough to guarantee the equilibrium is reached at all times.
We have to physically compress the gas. The reason behind this, along with answering your 2nd question, will be apparent if we analyze the Carnot cycle. It's convenient to consider the $PV$ and $TS$ diagrams when talking about processes:
Let's start at state 1 with $S_1$, $T_1$, $P_1$, etc.
1 $\rightarrow$ 2: Isentropic work input.
2nd law for this process gives:
$$ S_2 - S_1 = \int\frac{\delta Q_{12}}{T} + \Delta S_{gen}$$
Look at the TS diagram; clearly, $S_2 = S_1$. We also know $\Delta S_{gen} = 0$ since this is the Carnot cycle.
Therefore, $Q_{12} = 0$ and the 2nd law yields $S_1 = S_2$.
How else are we gonna increase the temperature to state 2? We have to add some sort non-heat energy to increase the temperature but keep $\Delta S = 0$. We therefore must add work.
2 $\rightarrow$ 3: Isothermal heat addition and work output.
The second law for this process gives:
$$ S_3 - S_2 = \frac{Q_{23}}{T_H}$$
where $T_H = T_2 = T_3$. Heat must therefore be added because we increase entropy. Note from the $PV$ diagram that we also have positive work output.
We've now made half a square on the $TS$ diagram. Next step is to isentropically decrease the temperature back to $T_1$.
3 $\rightarrow$ 4: Isentropic work output.
For the same reason as process 1 $\rightarrow$ 2, this process is adiabatic. The only way to decrease the entropy is to remove some sort of non-heat energy (i.e., the second law says we have to output work).
4 $\rightarrow$ 1: Isothermal heat rejection and work input.
The second law for this process gives:
$$ S_{1'} - S_4 = \frac{Q_{41'}}{T_C}$$
where $T_C = T_4 = T_1$. I'm using state $1'$ because we could reject heat $Q_{41'} = T_C(S_1' - S_4)$ that takes us to a different state. But, since this is a cycle, we reject just enough heat to bring us back to state $1$. We don't have to do that, we just choose to since this is a cycle. I think that answers your 2nd question.
Best Answer
This is exactly why there is energy exchange.
You should think in small steps: The volume is expanding, but the change occurs slowly enough to allow the system to continually adjust to the temperature of the reservoir through heat exchange.
See Isothermal process for more details.