Induced metric on $S^2$ from pullback of metric on $\mathbb{R}^3$ I'm going over some GR from more of a differential geometry perspective and had a quick question about a simple calculation - my differential geometry background isn't too strong so I apologise if any of the terminology is incorrect, but I'd be grateful for any clarification.
I'm following an example in Appendix A of Sean Carroll's Introduction to GR, where there is a map $\phi:M \to N$, given by
$$ \phi(\theta,\phi) = (\sin\theta \cos \phi , \sin \theta \sin \phi, \cos \phi) ,
$$
and $M=S^2$ is a submanifold of $N =\mathbb{R}^3$ (i.e. the two-sphere embedded in $\mathbb{R}^3$). The coordinates on the manifolds are $x^{\mu} = (\theta,\phi)$ on $M$, and $y^{\alpha} = (x,y,z)$ on $N$. The induced metric on $M$ is just the pullback of the flat-space metric $\phi^*g$, which is given by the formula
$$ 
(\phi^*g)_{\mu \nu} = \frac{\partial y^{\alpha} }{\partial x^{\mu}} \frac{\partial y^{\beta} }{\partial x^{\nu}} g_{\alpha \beta}.
$$
I understand how to calculate the individual Jacobian matrices of partial derivatives $\frac{\partial y^{\alpha} }{\partial x^{\mu}}$ (e.g. just using $y^1 = \sin\theta \cos \phi $, $y^2 = \sin \theta \sin \phi$ and $y^3 = \cos \phi$ as defined by $\phi$), however I was confused as to how to treat the full expression above.
The Jacobian matrices are $2 \times 3$ matrices, so the second has to be transposed to be a $3 \times 2$ matrix in order to give the required $2 \times 2$ metric $g_{\mu \nu}$. My question is, in the pullback equation above, how does the metric $g_{\alpha \beta}$ act on the Jacobian $\frac{\partial y^{\beta} }{\partial x^{\mu}}$, and how should I be writing this down? I can see that $y^{\beta}$ should be replaced with $y^{\alpha}$, but should any indices be lowered, and how should I interpret the metric tensor transposing the Jacobian matrix?

Edit - Ted Shriffin's comments are correct, the last component of the map should be $\cos \theta$ not $\cos \phi$, and all my matrices should be transposed.
 A: Seeing as I'm getting new comments on this old question I thought I'd post the answer myself. Thanks to Ted Shifrin's help in the comments and also pointing out the typo in the mapping: the map $\Phi:M \to N$ should of course be
$$ \Phi(\stackrel{x^{\mu} \ \rm{coords}}{\theta \, , \, \phi})^T = (\stackrel{y^\alpha \ \rm{coords}}{\sin\theta \cos \phi , \sin \theta \sin \phi, \cos \theta})^T \ ,
$$
where the transpose makes clear these are vectors not rows. I've also made clear the $x^{\mu}$ and $y^{\alpha}$ coordinates on $S^2$ and $\mathbb{R}^3$ respectively.
My mistake came from being unfamiliar with using matrix notation with tensors (it's been a while since I've done linear algebra). The final matrix of course must've been $2 \times 2$ because the indices $\alpha \beta$ were summed over whilst $\mu = \{0,1\}$. Similarly, all the expressions I had should have been transposed, but in the textbook they weren't written this way, probably to save space. e.g. the Jacobian was written as a $2 \times 3$ matrix but should have been $3 \times 2$,
$$
\frac{\partial y^\alpha}{\partial x^{\mu}} = 
\begin{pmatrix} 
\frac{\partial y^1}{\partial x^1} & \frac{\partial y^1}{\partial x^2}  \\
\frac{\partial y^2}{\partial x^1} & \frac{\partial y^2}{\partial x^2}   \\
\frac{\partial y^3}{\partial x^1} & \frac{\partial y^3}{\partial x^2}  
\end{pmatrix}
= 
\begin{pmatrix} 
\cos \theta \cos \phi & -\sin \theta \sin \phi \\
\cos \theta \sin \phi & \sin \theta \cos \phi  \\
-\sin \theta & 0 
\end{pmatrix} \ .
$$
Then the induced metric can be written as
$$ \begin{align}
(\Phi^* g)_{\mu \nu} &= \frac{\partial y^{\alpha}}{\partial x^{\mu}}  \frac{\partial y^{\beta}}{\partial x^{\nu}} g_{\alpha \beta} \\
&= \frac{\partial y^{\alpha}}{\partial x^{\mu}}  \frac{\partial y^{\beta}}{\partial x^{\nu}}
\delta_{\alpha \beta} \ ,
\end{align} 
$$
with the Kronecker delta from the flat Euclidean metric on $\mathbb{R}^3$. Here we don't just use matrix multiplication as I originally thought (this wouldn't make sense anyway). We can find the components of the induced metric by summing over the repeated index,
$$
(\Phi^* g)_{\mu \nu} = \sum_{\alpha = 1}^3 \frac{\partial y^{\alpha}}{\partial x^{\mu}}  \frac{\partial y^{\alpha}}{\partial x^{\nu}} .
$$
E.g. the $\mu \nu = 0 0$ component is just $$ 
(\cos \theta \cos \phi)^2 + (\cos \theta \sin \phi)^2 + (- \sin \theta)^2 = 1 \ ,
$$
and repeat for the others.
One then thankfully finds the induced metric to be
$$
g_{\mu \nu}(x) = \begin{pmatrix}
1 & 0 \\
0 & \sin^2 \theta \\
\end{pmatrix} \ .
$$
