# Is connectedness needed in this statement?

Suppose $$M$$ is a arcwise connected metric space, $$A_1$$ and $$A_2$$ are mutually separated in $$M$$. Let $$f_i:A_1\cup A_2\to M$$ for $$i=$$1,2 be two continuous functions such that $$f_0|A_i$$ ($$f_0$$ restricted to $$A_i$$) is homotopic to $$f_1|A_i$$ for $$i=$$1,2. Then $$f_0$$ is homotopic to $$f_1$$

• Here is the link about arcwise connectedness.
• Two sets A and B in a metric space are said to be mutually separated if They are disjoint and open in their union $$A\cup B$$
• Here is the link about homotopy

My question is

"Is arcwise connectedness really needed here to prove the above statement ?"

It seems arcwise connectedness is dispensable by below argument

Since $$f_0|A_i$$ homotopic to $$f_1|A_i$$, for $$i=1,2$$ There exists homotopy, $$F_i:A_i×[0,1]\to M$$ for $$i=1,2$$

Now define, $$F:A_1\cup A_2×[0,1]\to M$$ as $$F(x,t)= \begin{cases} F_1(x,t),& \text{if } x\in A_1\\ F_2(x,t), & otherwise \end{cases}$$

Here $$F$$ is well defined since $$A_1$$ and $$A_2$$ are disjoint. Also $$F$$ is a homotopy here irrespective of $$M$$ being arcwise connected.

Am I correct ?

• You do need the fact that $A_1$ and $A_2$ are separated to infer continuity of $F$, but apart from that your argument is right. Arcwise connectedness is not relevant here. – Rob Arthan Nov 8 '18 at 20:48

## 1 Answer

Yes, you are correct. The assumption that $$M$$ is arcwise connected is completely irrelevant (as is the assumption that $$A_1$$ and $$A_2$$ are subsets of $$M$$, and the assumption that $$M$$ is a metric space). The way I would state the result with no irrelevant hypotheses would be:

Let $$X$$ and $$M$$ be topological spaces and $$f_0,f_1:X\to M$$ be continuous. Suppose $$A_1$$ and $$A_2$$ are disjoint open subsets of $$X$$ whose union is $$X$$, and that $$f_0|_{A_i}$$ is homotopic to $$f_1|_{A_i}$$ for $$i=1,2$$. Then $$f_0$$ is homotopic to $$f_1$$.

• The $A_i$ don't need to be open in $X$, they just need to be open in $A_1 \cup A_2$. – Rob Arthan Nov 8 '18 at 20:50
• $X$ is just my name for $A_1\cup A_2$ (notice that I assume their union is $X$). – Eric Wofsey Nov 8 '18 at 20:52
• Yes, I agree, but I think you've thrown a little bit of the baby out with the bath-water. I'd guess it's a matter of taste. – Rob Arthan Nov 8 '18 at 20:54
• I think rather I've rescued the baby who was hidden and submerged in the bath-water. In typical applications, it's not that you start out with $A_1$ and $A_2$ and then observe that you get a homotopy on their union. Instead, you start with a space $X$ which can be decomposed into $A_1$ and $A_2$ and construct a homotopy separately on each one. – Eric Wofsey Nov 8 '18 at 20:56
• But I could counter that the "right" generalisation is to say: if $X$ and $M$ are topological spaces, if $A_1$ and $A_2$ are mutually separated subsets of $X$ and if $f_1: A_1 \to M$ and $f_2 : A_2 \to M$ are continuous functions, then $f_1$ and $f_2$ extend to a continuous function $f : A_1 \cup A_2 \to M$. (So I'm emphasising the criterion for the $A_i$ to be connected components of the subspace $A_1 \cup A_2$ and ignoring the "irrelevant' fact that the $f_i$ are homotopies, so my $A_i$ is $A_i \times [0, 1]$ in the original question.) The baby has too many aspects $\ddot{\smile}$. – Rob Arthan Nov 8 '18 at 22:15