Let $R$ be a commutative ring with $1$.
My professor defined $\operatorname{Tor}_i^R(M,N)$ as follows: Tensor a projective resolution $\dots\to P_1\to P_0\to M\to 0$ with $N$ and set $$\operatorname{Tor}_i^R(M,N):=\ker(P_i\otimes N \to P_{i-1}\otimes N)/\operatorname{im}(P_{i+1}\otimes N\to P_i\otimes N) \text,$$ where $P_{-1}:=0$.
To prove that $\operatorname{Tor}_0^R(M,N)=M\otimes_R N$, I have seen the following argument:
Choose a projective resolution $$\dots\to P_1\to P_0\xrightarrow\varepsilon M\to 0 \text;$$ this gives a short exact sequence $$0\to\ker\varepsilon\to P_0\xrightarrow\varepsilon M\to 0 \text.$$ Tensoring with $N$ yields the exact sequence $$0\to\ker\varepsilon\otimes N\to P_0\otimes N\xrightarrow{\varepsilon\otimes\operatorname{id}} M\otimes N\to 0 \text,$$ hence $$\operatorname{Tor}_0^R(M,N)=(P_0\otimes_R N)/(\ker\varepsilon\otimes N) \cong M\otimes N \text.$$
Now most of this is clear to me, except: In the sequence $$0\to \ker\varepsilon\to P_0\xrightarrow\varepsilon M\to0 \text,$$ why is $\ker\varepsilon$ projective? Is this even true, or is the definition incomplete and it does not need to be?
In general $\ker \varepsilon$ is not projective, otherwise every module would have a projective resolution of length one. Consider the following counter example: Let $R=K[x,y]$ where $K$ is a field and let $M=R/(x,y)$. Let $\epsilon: R \to M$ be the map that sends an element to its residue class. The kernel is the ideal $(x,y)$ which is not projective.
Also tensoring an exact sequence with $N$ does not give an exact sequence in general.