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Let $S=\{(x,a_3 , a_2, a_1 , a_0) \in \mathbb R^5 : x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 =0\}$
I want to show that $S$ is a connected manifold, and find the dimension of $S$.
It seems that each $x$ and $a_i$'s are all variables. If only $x$ is a variable, then $S$ cannot be connected.
I found that $S$ is trivially Hausdorff, but I have a hard time proving that $S$ is locally Euclidean and connected. How can I prove this?
The dimension comes directly from $U \cong \mathbb R^m$, then $m$ is the dimension.

bellcircle
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  • If you don't like how the variables are named, just rename them. You might want to use the regular value theorem for this one. See question: http://math.stackexchange.com/questions/913616/regular-value-theorem-using-implicit-function-theorem-in-calculus – b00n heT Jun 08 '16 at 07:57

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I'm who posted this question, and I finally found the solution. Look at the solution below.

Since $a_0 = -x^4 -a_3 x^3 -a_2 x^2 -a_1 x$, $S$ is a graph of a continuous function $g(x, a_3, a_2, a_1)=-x^4-a_3x^3-a_2x^2-a_1x$.
Define $f: \mathbb R^4 \to S $ as $f(x, a_3, a_2, a_1)=(x, a_3, a_2, a_1, -x^4-a_3x^3-a_2x^2-a_1x)$.
Then $f$ is continuous and bijective, and its inverse is the projection of $S$ onto $\mathbb R^4$, so the inverse is also continuous.
$\therefore f$ is a homeomorphism, so $S$ is homeomorphic to $\mathbb R^4$, and I concluded that $S$ is a 4-dimensional manifold.

Am I right?

bellcircle
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  • A solution fit perfectly to your case. Or use regular value theorem if you learn it as b00n heT said. – cjackal Jun 09 '16 at 14:26