No injective path between the two origins of the "line with two origins". During my preparations for my topology exam, I came across the following exercise:
Let L denote the line with two origins. Show that there is no injective path between the two origins $0_{-}$ and $0_{+}$. I tried to prove it by contradiction as follows:
I have already shown that $L$ is not Hausdorff, in particular, we cannot separate the two origins. So any subspace containing both origins cannot be Hausdorff. Assume that there exists an injective path $\gamma: [0,1] \to L$ from $0_{-}$ to $0_{+}$. Then $\gamma: [0,1] \to \gamma([0,1])$ is bijective and $\gamma([0,1])$ is a subspace of $L$ containing both origins.
Now I tried to show, that $\gamma: [0,1] \to \gamma([0,1])$ is actually a local homeomorphism which then would lead to a contradiction. However, I failed to do so. Is this the right approach for this exercise and how can I show that $\gamma^{-1}$ is continuous.
I somehow have the feeling that I miss some triviality...
 A: I don't see how we can show that $\gamma$ should be a local homeomorphism, but I have something else for you. The idea is that $\gamma$ cannot 'cross' the origin. $\gamma([0,1/2))$ and $\gamma([1/2,1])$ are connected, disjoint subsets of $L$. So one of them is contained in $(-\infty,0)$ and the other one is contained in $(0,\infty)$. Find a contradiction by examining the value of $\gamma(1/2)$, using the continuity of $\gamma$.
A: The injectivity will give a contradiction. Notice that if we subtract the two origins from $\gamma ([0,1])$, then we obtain a continuous injection from the connected set $(0,1)$ into the union of connected components $(-\infty ,0)\cup (0,\infty )$, so we have $\gamma (0,1)\subset (-\infty ,0)$ for example, but this will give a contradiction since this will mean that the set $(-\varepsilon ,0)$ for $\varepsilon $ sufficient close to $0$ will contain elements of $\gamma (0,1)$ both for $t$ close to $1$ and $t$ close to $0$, and connectedness implies that $(0,\delta )$ for some $\delta $ will be mapped into $(-\varepsilon ,0)$, so there must not be any elements of $\gamma (t)$ for $t$ close to $1$ in this neighbourhood.
