TITLE: Meaning of "efface" in "effaceable functor" and "injective effacement"
QUESTION [27 upvotes]: I'm reading Grothendieck's Tōhoku paper, and I was curious about the reasoning behind the terms "effaceable functor" and "injective effacement". I know that in English, to efface something means to erase it, but I'm not sure if there's another meaning in either conversational or mathematical French. 
Given an abelian category $C$ and an additive category $D$, an effaceable functor $F:C\rightarrow D$ is an additive functor such that for all $A\in\text{Ob}(C)$, there exists some monomorphism $u:A\rightarrow M$ such that $F(u)=0$. I'm guessing that the "effacing" going on here is the functor $F$ killing the map $u$.
What I'm less sure about is the reasoning behind "injective effacement". Given an abelian category $C$ and $A\in\text{Ob}(C)$, an injective effacement of $A$ is a monomorphism $f:A\rightarrow M$ such that for any monomorphism $g:B\rightarrow C$, and any map $h:B\rightarrow A$, there is a map $t:C\rightarrow M$ such that the diagram commutes:
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{lll}
B & \ra{g} & C  \\
\da{h} & & \da{t} \\
A & \ra{f} & M \\
\end{array}
$$
Grothendieck points out that $\text{id}_M:M\rightarrow M$ is an injective effacement of $M$ if and only if $M$ is an injective object of $C$, and that any monomorphism of $A$ to an injective object $M$ is an injective effacement of $A$. I don't really see what we're erasing here; my guess is that an injective effacement $f:A\rightarrow M$ is somehow "erasing" the cohomology of $A$, but this is only a vague sense.

REPLY [13 votes]: Theo's answer is excellent. Let me add a bit on how this formalism is used in practice.
Say you want to show that $\mathrm{Tor}$ commutes with flat base change of rings. That is, if $M, N$ are $A$-modules, $B$ a flat $A$-algebra, then $\mathrm{Tor}_B(M_B, N_B) \simeq \mathrm{Tor}_A(M,N) \otimes_A B$ functorially -- it's not even obvious a priori how to get the map. The point is that both are universal $\delta$-functors in either variable, though, and since there is a clear isomorphism in dimension zero, it follows that the two functors are not only isomorphic, but isomorphic as $\delta$-functors. (To see that they are co-effaceable, one uses the fact that base change preserves projectivity, for instance.)
In Hartshorne, for instance, there are oodles of applications of this formalism, e.g. in showing that sheaf cohomology is the same derived functor whether you work in the category of all sheaves or just sheaves of modules over some sheaf of rings (the point is that either way, the derived functors are effaceable by flasque sheaves, which mean the same in either category). This last fact can also be seen if you accept that derived functors can be computed using acyclic (rather than injective) resolutions.
Let me add one more example, which is perhaps more interesting. Suppose $A$ is a noetherian ring and $I \subset A$ an ideal. Then we can consider the functor on the category of $A$-modules that sends an $M$ to the submodule consisting of all $m \in M$ killed by a high power of $I$. This is a left-exact functor, and its derived functors are the local cohomology modules $H^i_I(M)$.
It turns out that $H^i_I(M) \simeq \varinjlim_n \mathrm{Ext}^i(R/I^n, M)$ functorially; a quick way of seeing this is that it is clearly true in dimension zero (by inspection), and for higher dimensions we see that both are effaceable $\delta$-functors in $M$ (since filtered colimits are exact for modules).