# Question about definition of $\mathrm{Ext}$

One can define $\mathrm{Ext}^n(M,N)$ (where $M,N$ are $R$-modules) in two ways, either by taking an injective resolution of $N$ and applying $\mathrm{Hom}(M,-)$or by taking a projective resolution and applying $\mathrm{Hom}(-,N)$.

I'm struggling to remember which is which and I'm sure there is a(n obvious?) reason. Why does it have to be projective + $\mathrm{Hom}(-,N)$ or injective + $\mathrm{Hom}(M,-)$ but cannot be projective + $\mathrm{Hom}(M,-)$?

-
Which one has the lifting property in which direction? – Qiaochu Yuan Jul 18 '12 at 18:09
Always use injective resolutions for right derived functors. (The fact that you can use a projective resolution here is because there's an ${}^\textrm{op}$ floating around!) – Zhen Lin Jul 18 '12 at 18:11
The projectives are the injectives of the category $(\text{$R$-$\mathrm{Mod}$})^{\mathrm{op}}$. Eventually, you will just remember which to use when, independently of these rationalizations :) – Mariano Suárez-Alvarez Jul 18 '12 at 18:13
@MattN. You say your definition of $\mathrm{Ext}$ uses projective resolutions. What Mariano and Zhen said is that, due to the "contravariant" nature of $\mathrm{Hom}(-,N)$, your projective resolution is really an injective resolution, but in the opposite category. All that says is that in both cases you're really doing the same thing. – M Turgeon Jul 18 '12 at 19:13
@MattN. By definition, projective modules are the ones that have nice properties when you look at functions out of them, and injectives are the ones that have nice properties when you look at functions into them. That's one way to remember which side to use. – Tyler Lawson Jul 18 '12 at 20:34