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 $$f_1$$

• Two sets A and B in a metric space are said to be mutually separated if They are disjoint and open in their union $$A\cup B$$

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

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
• 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