# Suppose that $f:[0,1]\to M$ is locally injective. Show that $f$ is piecewise injective.

Consider a metric space $$M$$ and $$f: [0,1]\to M$$ locally injective, i.e. for each $$x\in M$$ there is a neighborhood on which $$f$$ is injective. Show that $$f$$ is piecewise injective, i.e. there exists $$0= t_0 < t_1 < \dots < t_N = 1$$ such that $$f$$ is injective on $$[t_{n-1},t_n]$$ for each $$n\in\{1,\dots, N\}$$.

My attempt:

Given is that $$\forall x\in M: \exists \delta_x>0: \forall y,z\in ]x-\delta_x,x+\delta_x[: f(y)=f(z)\Rightarrow y=z$$. Then $$[0,1]\subseteq\bigcup_{x\in [0,1]} ]x-\delta_x,x+\delta_x[ \quad \Rightarrow \quad [0,1]\subseteq\,\, ]x_1-\delta_{x_1},x_1+\delta_{x_1}[\,\, \cup\dots\cup \,\, ]x_N-\delta_{x_N},x_N+\delta_{x_N}[,$$ for some $$x_1,\dots,x_N\in [0,1]$$. Assuming w.l.o.g. that $$x_1< \dots < x_N$$, then $$0\in \,\,]x_1-\delta_{x_1},x_1+\delta_{x_1}[\,\, =: B(x_1,\delta_{x_1})$$ and $$1\in B(x_N,\delta_{x_N})$$ (I'm not too sure about this). I believe that all the open intervals that cover $$[0,1]$$ will have to overlap (otherwise, i.e. no overlapping intervals or at least two intervals that don't overlap, we won't cover whole $$[0,1]$$). By 'all' I mean that for each interval $$B(x_i,\delta_{x_i})$$ we can find $$B(x_j,\delta_{x_j})$$ such that $$\exists t_{ij}\in B(x_i,\delta_{x_i})\cap B(x_j,\delta_{x_j})$$. W.l.o.g. we can assume that $$B(x_i, \delta_{x_i})$$ intersects $$B(x_{i+1}, \delta_{x_{i+1}})$$ for $$i=1,\dots, N-1$$. Then we can select $$t_{i-1}\in B(x_i,\delta_{x_i}), i=1,\dots, N$$.

Is this somewhat correct?

You have a finite set of intervals $$(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)$$ that cover $$[0,1]$$ and are such that $$0\in(x_1,y_1)$$, and $$1\in(x_n,y_n)$$. You can further assume that none of these intervals is a subset of another: if one were, you could simply throw it away.
Let $$(u_1,v_1)=(x_1,y_1)$$. If $$v_1>1$$, then $$[0,1]\subseteq(u_1,v_1)=(x_1,y_1)$$, and $$f$$ in injective. Otherwise, there is an interval $$(x_k,y_k)$$ that contains $$v_1$$, and we let $$(u_2,v_2)=(x_k,y_k)$$. Proceeding in this fashion, in a finite number of steps we reduce the open cover to a family $$\{(u_1,v_1),\ldots,(u_m,v_m)\}$$ such that $$0\in(u_1,v_1)$$, $$1\in (u_m,v_m)$$, and $$v_k\in(u_{k+1},v_{k+1})$$ for $$k=1,\ldots,m-1$$.
Now you have a cover of $$[0,1]$$ by open intervals that overlap in a fairly predictable way, and you can define the desired partition: let $$t_0=0$$, $$t_k\in(u_{k+1},v_k)$$ for $$k=1,\ldots,m-1$$, and $$t_m=1$$. Then $$[t_{k-1},t_k]\subseteq(u_k,v_k)$$ for $$k=1,\ldots,m$$, so $$f$$ is injective on each segment of the partition.