For my topology class I have to show the following:
Let $g = g_{ij} \; dx^i \otimes dx^j$ be a metric on a differentiable manifold. Show that the volume form $dV = \sqrt{\det g} \; dx^1 \land ... \land dx^n$ is invariant under coordinate transformations.
Now, I am not quite sure what I have to do here. If $x = (x^1, ..., x^n) $ is a set of coordinates and $\tilde{x} = (\tilde{x}^1, ..., \tilde{x}^n)$ is another set of coordinates, then do I have to show that
$$d\tilde{V}( \tilde{x} ) \enspace = \enspace dV(x) \quad ?$$
But how do I know how $d\tilde{V}(\tilde{x})$ looks like? Does anyone have a hint for me?
Using basic coordinate transformation rules for tensors ($g_{ij}$ and $dx^i$), as well as antisymmetric properties of the wedge product, show that the wedge product
$dx^1\wedge\dots dx^n$ transforms as follows under coordinate transformations:
$$ dx^1\wedge\dots dx^n=\frac{\partial(x)}{\partial(\tilde{x})}\,d\tilde{x}^1\wedge\dots d\tilde{x}^n $$
Where $\frac{\partial(x)}{\partial(\tilde{x})}$ is the determinant of the Jacobian matrix. Then, again using coordinate transformations, show that
$$ \mbox{det}\left[\tilde{g}\right]=\left(\frac{\partial(x)}{\partial(\tilde{x})}\right)^2\,\mbox{det}\left[g\right] $$
Conclude that, under coordinate transformations:
$$ dx^1\wedge\dots dx^n=\sqrt{\frac{\mbox{det}\left[\tilde{g}\right]}{\mbox{det}\left[g\right]}}\frac{\partial(x)}{\partial(\tilde{x})}\,d\tilde{x}^1\wedge\dots d\tilde{x}^n $$
Finally conclude that:
$$ \sqrt{\mbox{det}\left[g\right]}\cdot dx^1\wedge\dots dx^n = \sqrt{\mbox{det}\left[\tilde{g}\right]}\cdot d\tilde{x}^1\wedge\dots d\tilde{x}^n $$
i.e. the form of this quantity does not change as you change coordinates.
Finally, finally, note that any manifold is locally a Cartesian space, and there, in Cartesian coordinates $\mbox{det}\left[g\right]=1$ and $dx^1\wedge \dots dx^n=d^nV$, the standard volume form. From this conclude that $d^n V=\sqrt{\mbox{det}\left[g\right]}dx^1\wedge \dots dx^n$ is the correct volume form in any coordinate system and at all points on the manifold
Adding more specifics. I would suggest https://www.amazon.com/Tensors-Differential-Variational-Principles-Mathematics-ebook/dp/B00A735HK8#reader_B00A735HK8. It deals with all that you want in a rigorous and hands-on-way.
To compute the determinant of the metric tensor you need a way of expressing this determinant using tensors. At this point on usually reaches for Levi-Civita, but the problem with that is that Levi-Civita is not a tensor (it is a relative tensor, in the language of Lovelock and Rund). Fortunately, there is something called generalized Kroenecker delta [https://en.wikipedia.org/wiki/Kronecker_delta#Definitions_of_the_generalized_Kronecker_delta], which is a tensor (to prove this note that generalized Kroenecker delta can be written as products of usual Kronecker delta's $\delta^{\mu}_{\nu}$, so it is sufficient to prove that latter is a tensor).
Ok, with that:
$$ \det\left[\tilde{g}\right]=\tilde{\delta}^{\alpha_1\dots\alpha_n}_{1\dots n}\tilde{g}_{1\alpha_1}\dots\tilde{g}_{n\alpha_n} $$
To check that, note that $\tilde{\delta}^{1\dots n}_{1\dots n}=1$ and then picks up a factor of -1 on every exchange of $\alpha$-s. Now you can apply change of coordinates to the RHS, since there you only have tensors:
$$ \begin{align} \det\left[\tilde{g}\right]=\frac{\partial x^{\mu_1}}{\partial \tilde{x}^1}\dots\frac{\partial x^{\mu_n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1\dots\alpha_n}_{\mu_1 \dots \mu_n} \cdot \frac{\partial x^{\sigma_1}}{\partial \tilde{x}^1}\dots \frac{\partial x^{\sigma_n}}{\partial \tilde{x}^n} \cdot g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n} \end{align} $$
Firstly deal with sum over $\mu$-s. Let's look at just two terms, for example:
$$ \begin{align} \frac{\partial x^{\mu_1}}{\partial \tilde{x}^1}\dots\frac{\partial x^{\mu_n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1\dots\alpha_n}_{\mu_1 \dots \mu_n} =& \frac{\partial x^{1}}{\partial \tilde{x}^1}\frac{\partial x^{2}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n} + \frac{\partial x^{2}}{\partial \tilde{x}^1}\frac{\partial x^{1}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1 \dots\alpha_n}_{2, 1\dots n}+\dots\\ =& \frac{\partial x^{1}}{\partial \tilde{x}^1}\frac{\partial x^{2}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n} - \frac{\partial x^{2}}{\partial \tilde{x}^1}\frac{\partial x^{1}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n}+\dots \end{align} $$
Note the swap of indices on the least delta. In this way we can show that the sum over $\mu$-s consist of a product $\delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n}$ and a completely anti-symmetrized product of partial derivatives - a Jacobian determinant:
$$ \begin{align} \det\left[\tilde{g}\right]=\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\cdot \delta^{\alpha_1\dots\alpha_n}_{1 \dots n} \cdot \frac{\partial x^{\sigma_1}}{\partial \tilde{x}^1}\dots \frac{\partial x^{\sigma_n}}{\partial \tilde{x}^n} \cdot g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n} \end{align} $$
Next consider:
$$ w_{\sigma_1\dots\sigma_n}=\delta^{\alpha_1\dots\alpha_n}_{1 \dots n} g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n} $$
And note that:
$$ w_{1\dots n}=det\left[g\right] $$
and any exchange of $\sigma$ indices simply adds a factor of -1 to $w_{\sigma_1\dots\sigma_n}$. We can therefore repeat the trick above to get
$$ \begin{align} \det\left[\tilde{g}\right]=&\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\cdot w_{1\dots n} \cdot \frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\\ =&\left(\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\right)^2\cdot\det\left[g\right] \end{align} $$