Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 2

A topological space is connected if it cannot be expressed as a union of two disjoint nonempty sets open in that space. Suppose $f(X)$ had such a separation $U\cup V$. Here $U,V$ are open in the subspace topology of $f(X)$. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$. They are disjoint, nonempty because $U,V$ contain image points, and cover $X$. This contradicts connectedness. Hence a continuous image of a connected space is connected.

A space is path-connected if for every pair $x_0,x_1$ it contains a continuous path $\gamma :[0,1]\to X$ with $\gamma (0)=x_0$ and $\gamma (1)=x_1$. For any $y_0,y_1\in f(X)$ choose preimages $x_0,x_1$. A path between those preimages maps to $f\circ \gamma$, a continuous path from $y_0$ to $y_1$ staying in $f(X)$. Thus a continuous image of a path-connected space is path-connected. Surjectivity onto the ambient codomain $Y$ is not needed; the conclusion concerns $f(X)$.

Suppose a path-connected space had a separation $X=U\cup V$ as above. Choose $u\in U$, $v\in V$, and a continuous path $\gamma$ joining them. The nonempty disjoint sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$ would separate $[0,1]$, contradicting connectedness of an interval. Hence path-connectedness implies connectedness. Equivalently, the connected path image from part (i) cannot meet both sides of a separation.

Expanding the characteristic polynomial givesTherefore $a=-2z$ and $b=z^2-x^2-y^2$. Both are polynomial functions of the three real coordinates, so they are continuous functions, including on the open subset $X$ of invertible matrices.

On $X$, $b=\det A\ne 0$. Its image contains positive values, for example $b(I)=1$, and negative ones, for example $b(\mathrm{diag}(1,-1))=-1$. If $X$ were connected, its continuous real image would be an interval containing both signs and hence zero. That is impossible. Thus $X$ is not connected. More precisely its determinant image is $\mathbb{R}\setminus \{0\}$: take $x=y=0$ for a positive value and $z=y=0$ for a negative one.

On $X^{+}$, $z^2>x^2+y^2$, so $z\ne 0$ and therefore $a=-2z\ne 0$. Yet $a(I)=-2$ and $a(-I)=2$. The same continuous image of a connected space argument shows that $X^{+}$ is not connected.

Put $r=\sqrt{x^2+y^2}$. The required three disjoint nonempty open sets areTheir union is precisely $X$, the complement of the double cone $z^2=x^2+y^2$. The eigenvalues are $z\pm r$, so $X_1$ consists of positive-definite matrices above the upper sheet, $X_2$ of negative-definite matrices below the lower sheet, and $X_3$ of indefinite matrices outside the double cone. Examples are $I,-I,\mathrm{diag}(1,-1)$ respectively.

The two definite cones are convex sets. The indefinite region is path-connected: shrink $z$ continuously to zero while keeping $r>0$, then change the nonzero vector $(x,y)$ within the punctured plane. Thus these sets are also exactly the three connected components. This is the determinant cone for real symmetric two-by-two matrices.