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Consider the subspace of $\mathbb{C}\times\mathbb{C}$: \begin{equation} Y = \{(w,z) \in \mathbb{C}\times\mathbb{C}| w^2 = z^4 - 1\} \end{equation}

The goal is to show that it is path-connected and is non-compact. How would one go about showing those properties? (I find it hard to visualize a subspace of $\mathbb{C}\times\mathbb{C}$ and so I don't exactly how an intuition of how to start.)

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    @JoeJohnson126: This is not the graph of a continuous function, so I'm not sure what you had in mind with your first comment. – 2017-01-13
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    Well if we put $f(w,z)=w^2-z^4+1$, then $Y=f^{-1}(\{0\})$, so you know $Y$ is closed. So to show it is not compact, you need to show that $Y$ is not bounded. – 2017-01-13
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    @Aweygan Is it necessary to say that $Y$ is closed? Since Heine Borel is only application for $\mathbb{R}^n$, we don't need the closed and bounded = compact condition right? For a metric space compact = complete and totally bounded and totally bounded $\Rightarrow$ bounded. Hence, we only need to show $Y$ is not bounded. – 2017-01-13
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    @UnrealAnalysis $\mathbb C ^n$ is isometric to $\mathbb R ^{2n}$, so Heine-Borel still applies. Nevertheless, your approach to not-bounded $\implies$ not-compact is also valid – 2017-01-13

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Here's a fuller picture, assuming that you, like I, cannot visualize 4 dimensions.

What you can visualize are two separate complex planes, the $w$-plane and the $z$-plane.

A path in $Y$ is equivalent to a pair of paths $w(t)$ in the $w$ plane and $z(t)$ in the $z$ plane, $t \in [0,1]$, subject to the constraint that $$w(t)^2 + 1 = z(t)^4 $$ You can use path lifting theorems from topology to study this situation. Here's an outline.

The point where $z=0$ has two corresponding $w$ values, $w=\pm i$, giving two points $(z,w) = (0,\pm i)$ in $Y$. Also, the point where $w=0$ has four corresponding $z$ values, $z=\pm 1, \pm i$, giving four points $(z,w) = (\pm 1,0), (\pm i,0)$. That gives 6 special points in $Y$, a subset I'll denote $Q \subset Y$.

If we let $p_w : Y \to \mathbb{C}$ be the projection to the $w$ plane, and if we let $Q_w = p_w(Q) = \{\pm i, 0\}$, then the map $p_w : Y - Q \to \mathbb{C} - Q_w$ is a degree 4 covering map.

Similarly, if we let $p_z : Y \to \mathbb{C}$ be the projection to the $z$ plane, and if we let $Q_z = p_z(Q) = \{0,\pm 1, \pm i\}$, then the map $p_z : Y - Q \to \mathbb{C} - Q_z$ is a degree 2 covering map.

Any path $\gamma(t)=(w(t),z(t))$ in the punctured surface $Y-Q$ can be thought of as a lift via $p_z$ of the path $z(t)$, or as a lift via $p_w$ of the path $w(t)$.

So, pick some base point in $Y-Q$, say $(z_0,w_0) = (2,\sqrt{15})$. Pick any other point $(z_1,w_1) \in Y-Q$. Imagine a path $w(t)$ that starts at $w_0=\sqrt{15}$ and wanders around in $\mathbb{C}-Q_z = \mathbb{C}-\{0,\pm 1,\pm i \}$; using path connectivity you can clearly get to $w_1$. Lift the path $w(t)$ up to $Y-Q$ and project down to a path $z(t)$ $\mathbb{C}-Q_z$. You might get lucky and hit $z_1$, but because the map $p_z : Y-Q \to \mathbb{C}-Q_z$ is 2-to-1 there is one other possibility I'll denote $z'_1$.

So, can you find a closed path in $\mathbb{C}-Q_z$ from $w_1$ to $w_1$ whose lift to $Y-Q$ projects to a path in $\mathbb{C}-Q_z$ from $z_1$ to $z'_1$?

If so, you win, you've proved $Y-Q$ is path connected and so $Y$ is path connected.

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    Should $p_w$ be degree 4 and $p_z$ be degree 2 instead? If I understand the degree of a covering map correctly: for each point $w \in \mathbb{C} - Q_w$, there are 4 possible $z$. Hence, there are 4 possible points in $Y - Q$ and hence $p_w$ is degree 4. – 2017-01-14
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    Oops, you're right about that, I'll fix it, which changes the roles of $z$ and $w$ in the latter part of my answer. – 2017-01-14
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    Currently, the main difficulty is that although we can easily find a path in $\mathcal{C} - Q_w$ from $w_0$ to $w_1$, $z_0$ might not hit $z_1$. I want to show that it is always possible to hit $z_1$. It suffices to show that there is always a path between $(z_0, w_0)$ and $(z_1, w_0)$. Is there some theorem in path lifting theorem that allows us to show that? – 2017-01-15
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    You need just one closed loop in $\mathbb{C}-Q_w$ whose lift is not closed. Testing generating loops of the fundamental group of $\mathbb{C}-Q_w$ is sufficient. – 2017-01-16
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Here's a useful observation: for any $w \in \Bbb C$, there exists at most four values of $z$ (and at least one) such that $(z,w) \in Y$. This alone is enough to conclude that $Y$ is unbounded.

Similarly, for any $z \in \Bbb C$, there exists at most two (and least one) value of $w$. Note also that for $(w,z) \in Y$, $w^2 = -1 \iff z = 0$.

Here's a strategy for proving path connectedness: for any $(w,z)$ with $z \neq 0$, show that we can slide the value of $z$ along in $Y$ to get to one of the points $(\pm i, 0)$. Then, show that there is a path between the points $(\pm i, 0)$.

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    Would the function $f : [0,1] \mapsto \mathbb{C}\times\mathbb{C}$ defined by $f(t) = \big(e^{\frac{1}{2}\log{t^4z^4-1}}, tz\big)$ work? $f$ is clearly continuous for the second component. For the first, might there be a problem if the term in the log is 0? – 2017-01-13
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    The complex logarithm is not well-defined, so it is usually a bad idea to use it in this fashion, unless you are very careful about restricting its domain and range in such a way to make it well defined (similarly to how the real-valued square root is not well-defined, unless you restrict the domain and range to the positive numbers). – 2017-01-13
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    @LeeMosher I see. Do you have a suggestion on how $f(t)$ can be defined? – 2017-01-13