Proving that this space is path connected.

If $X$ is a topological space, then I have to show that $(X \times D^1)/$~ is path connected (where $(x_1, y_1)$ ~ $(x_2, y_2)$ $\iff$ $y_1=y_2=1$ or $y_1=y_2=-1$ or $(x_1, y_1) = (x_2, y_2)$ ).

Let $(x_1, y_1), (x_2,y_2) \in (X \times D^1)/$~. Then we can draw a line connecting $(x_1, y_1)$ with $(x_1, 1)$. Because $(x_1,1)$ ~ $(x_2, 1)$. We can then extend that line from $(x_2, 1)$ to $(x_2, y_2)$. So we can just define the function $f: [0,1] \rightarrow (X \times D^1)/$~ with that line such that $f(0)=(x_1,y_1)$ and $f(1)=(x_2,y_2)$.

Is my answer correct? I feel like its too simple, so I tried to actually construct an explicit function but wasn't able to...so I was wondering if anybody could give me a hint (if my answer is not rigorous enough).

-

The formula given by rewritten is correct. Although there is no addition in $X$, the addition in $f$ is only happening in the $D^1$ component, for which addition is well-defined (as a subset of $\mathbb{R}$: notice that if $y \in D^1$ and $0 \leq t \leq 1/2$, $1/2 \leq t' \leq 1$, then $(2t)(1-y), 2(1-t')(1-y)$ are again in $D^1$). Thus $f$ still makes sense in the quotient, and defines a valid path from $(x_1, y_1)$ to $(x_2, y_2)$.

EDIT: Here is a complete answer:

As in your comment below, given points $(x_1,y_1), (x_2,y_2) \in X \times D^1 / \! \sim$, define the maps \begin{align*} &f_1 : [0,1/2] \to X, \qquad f_1(t) = x_1 \\ &f_2 : [0,1/2] \to D^1, \qquad f_2(t) = (2t)(1-y_1) + y_1 \\ &f_1' : [1/2,1] \to X, \qquad f_1'(t) = x_2 \\ &f_2' : [1/2,1] \to D^1, \qquad f_2(t) = 2(1-t)(1-y_2) + y_2 \end{align*}

$f_1, f_1'$ are continuous because they are constant maps (if this is not clear you should check it, using any definition of continuity you wish - the reasoning you gave is not quite enough). $f_2, f_2'$ are continuous because addition and multiplication are continuous on $\mathbb{R}$ (or in any case, $f_2, f_2'$ are linear functions in $t$, and continuity can be checked easily - think of the graphs). Thus, by characterization of the product, we have continuous maps

$$(f_1, f_2) : [0,1/2] \to X \times D^1,$$ $$(f_1', f_2') : [1/2,1] \to X \times D^1$$

Composing each of these with the (continuous) quotient map $p : X \times D^1 \to X \times D^1 / \! \sim$, we get maps $p \circ (f_1, f_2) : [0,1/2] \to X \times D^1 / \! \sim$, $p \circ (f_1', f_2') : [1/2,1] \to X \times D^1 / \! \sim$. These paste together to give a single map on $[0,1]$ iff they agree on the intersection, which is just $\{1/2\}$. But at $t = 1/2$, the first map gives (the class of) $(x_1,1)$, and the second gives (the class of) $(x_2,1)$, and these are indeed the same in the quotient.

A few notes:
i) Note carefully the ranges of the various maps (e.g., $f_2$ does not go into the quotient).
ii) It is not true that the closure of any subset $A \subseteq [0,1/2]$ is of the form $[a,b]$. Look up the Cantor set for an extreme example. The reasoning you gave for continuity of $f_2$ is thus not correct.
iii) If you're still having trouble seeing why $f_2, f_2'$ are continuous, try proving that any linear function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = mx + b$ is continuous, e.g. by the $\epsilon$-$\delta$ definition.

-
I also thought so (since there is addition in $D^1$), but I'm not sure why my professor said it doesn't make sense...also, do you mind if I write my proof about the continuity of this function so you can check if its right or wrong? – user58289 Dec 14 '13 at 15:31
Sure, go ahead. – zcn Dec 14 '13 at 20:03
Let $f_1 : [0, 1/2] \rightarrow \sum X$ with $f_1(t)=x_1$. Let $f_2: [0,1/2] \rightarrow \sum X$ with $f_2(t) = (2t)(1-y_1)+y_1$. Let $f'_1: [1/2, 1] \rightarrow \sum X$ with $f'_1(t)=x_2$. And, finally, let $f'_2: [1/2] \rightarrow \sum X$ with $f'_2(t)=(2y_2-2)t + 2 - y_2$. So basically define the piecewise function $f$ by $f(t) = (f_1, f_2)$ for $t \leq 1/2$ and $f(t) = (f'_1, f'_2)$ for $t \geq 1/2$. In order to decide if $f$ is continuous, we need to check if $(f_1, f_2)$ and $(f'_1, f'_2)$ are continuous. – user58289 Dec 15 '13 at 17:16
$(f_1, f_2)$ is cont. if and only if $f_1$ and $f_2$ are. $f_1$ is cont. because the preimage of $\{x_1 \}$ is the whole domain. So if $\{x_1\}$ is open in $X$, then its preimage is also open (since its the whole thing). Also, if $\{x_1\}$ is closed then so is its preimage. So $f_1$ must be continuous. For $f_2$, for any subset $A \subseteq [0,1/2]$, we have that the closure of $A$ is of the form $[a,b]$. So $f_2(\bar{A}) = [(2a)(1-y_1)+y_1,(2b)(1-y_1)+y_1]=\overline{f_2(A)}$. Since $f_2(\bar{A}) \subseteq \overline{f(A)}$, $f_2$ is cont. The proof for $(f'_1, f'_2)$ is similar, right? – user58289 Dec 15 '13 at 17:27
Your reasoning is not quite correct - I've added a full answer – zcn Dec 16 '13 at 1:42

The answer is correct (actually you don't even need the $y_1=y_2=−1$ part for it to be path connected), you can construct a path between to points $(x_1, y_1)$ and $(x_2, y_2)$ by pieces:

$$f(t) = (x_1, y_1+(2t)(1-y_1)) \text{ if } 2t\leq1$$

$$f(t) = (x_2, y_2+(2(1-t))(1-y_2)) \text{ if } 2t\geq1$$

They are linear in $y$, constant in $x$, and they coincide for $t=\frac{1}{2}$.

-
Thanks a lot, but the second one should be $f(t) = (x_2, y_2 + (2(1-t))(1-y_2)-y_2)$, right? – user58289 Dec 4 '13 at 13:03
I don't think so. For $t=\frac{1}{2}^+$, $f(t) = (x_2, y_2 + (2(1-\frac{1}{2}))(1-y_2)) = (x_2, 1)$ and for $t=1$, $f(t) = (x_2, y_2+(2(1-1))(1-y_2)) = (x_2, y_2)$. – rewritten Dec 5 '13 at 11:59