What is the work done against a force? Suppose a particle travels a path $\gamma : I\subset \mathbb{R}\to \mathbb{R}^3$ subject to a force $\mathbf{F}: \mathbb{R}^3\to T\mathbb{R}^3$, then we know that we define the work done by the force as
$$W=\int_\gamma \mathbf{F} = \int_I \langle \mathbf{F}\circ \gamma, \gamma'\rangle$$
Now usually I see the term "work done against a force" and I don't really understand what it means. The reason is that in my understanding, work is always done by a force upon a particle or system of particles. If we talk about work done against a force, it is work done by which force on which particle or system?
Also, mathematically, how it is obtained? If we want to know not the work done by a force, but against it how we obtain it? I imagine is just the opposite, just changing the sign, but I'm unsure, and if it is really that I can't grasp why it should be.
 A: Note that the notion of work $W=\int_{\gamma} {\bf F}\cdot \mathrm{d}{\bf r}$ depends on the force ${\bf F}$. Here ${\bf F}$ could e.g. be a gravitational force or a frictional force, etc., leading to a gravitational work or a frictional work, respectively. The phrase 

work done against a force 

is shorthand for

work $W_1$ done by a force ${\bf F}_1$ against another force ${\bf F}_2$. 

A: Work is done by something or someone, expending energy by exerting a force on an object. I'm not quite sure what the context is of your quote, but I'd imagine that it could be a scenario much like the following one:
Imagine a ball with some mass $m$ being lifted by a man. The gravitational force from Earth is pulling the ball down, but the person nonetheless manages to lift the ball upwards. He expends an amount of work $W\geq mg\Delta h$ where $\Delta h$ is the difference between the initial and final height of the ball. He has now done work against the gravitational force. I think you might be simply reading too much into it all.
A: Assume you have a force field (it can be caused by eletric, magnetic, whatever fields). So, if you have a object in position $\vec{r}$, it will feel a force $\vec{F}(\vec{r})$.
So.. inside the vector field, now comes the question: What's the work I have to do to move a particle from here to there? Or, equivalently: What's the work I have todo against the force field, in order to move the particle from $A$ to $B$?
Mathematically, we do often talk about "quasi-static" movement. This means, the motion is so slow, soooo static, that it's acceleration $\vec{a} = \vec{0}$ and it's velocity $\vec{v}\to\vec{0}$. Since the movement is quasi-static, the force I have to apply $\vec{F'}$ in each position is just the opposite of the force field. So, the work I have to do against the force in a quasi-static movement:
$$
W = \int_\gamma \vec{F'}\cdot d\vec{r} = -\int_\gamma \vec{F}\cdot d\vec{r}
$$
Because $\vec{F'} = -\vec{F}$, since the motion is quasi-static. For this, the net force will give: $\vec{F'} + \vec{F} = \vec{0}$. And then we see the acceleration will indeed be zero. 
Note: That's precisily the definition of the potential energy $U$ if force field is conservative. We often calculate: $\vec{\nabla}U = -\vec{F}$. Because:
$$
\Delta U = \int_\gamma \vec{F'}\cdot d\vec{r} = 
-\int_\gamma \vec{F}\cdot d\vec{r} =
\int_\gamma \vec{\nabla}U\cdot d\vec{r} = 
U(B) - U(A).
$$
Then, potential energy can be defined as the work against a conservative force field to move particle from $A$ to $B$ in quasi-static motion.
