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It is said that the empty set is the initial object of the category of sets, and the empty space is the initial object in the category of topological spaces. So there exists a unique arrow from the initial object to any object in those categories. But I wonder what would be that map?

It seems that at $t = 0$ the map would give an empty set or space, while at $t = 1$ the map would give a non-empty set or space. How is that possible with a continuous map?

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  • $\begingroup$ What is $t=0, t=1$ ? What does a simple continuous map have to do with homotopies ? $\endgroup$ – Max Sep 5 '18 at 9:12
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    $\begingroup$ A homotopy $X\to Y$ can be looked at as a continuous function $F:X\times\mathbb I\to Y$. In special case $X=\varnothing$ then $X\times\mathbb I=\varnothing$ and the empty function is the only homotopy $\varnothing\to Y$. $\endgroup$ – drhab Sep 5 '18 at 9:13
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    $\begingroup$ The homotopy does not go from X to Y, rather from one such map to another such map. $\endgroup$ – Leon Hendrian Sep 5 '18 at 12:47
  • $\begingroup$ Yes, @LeonHendrian , as long as X is empty and Y is non-empty, is what I need. $\endgroup$ – Yan King Yin Sep 6 '18 at 13:14
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    $\begingroup$ @YanKingYin A function $f:A\to B$ is a set of ordered pairs with the special property that for every $a\in A$ there is a unique $b\in B$ with $(a,b)\in f$. The set $\varnothing$ is subset of $\varnothing\times B$. Further vacuously all elements of $\varnothing$ are ordered pairs (no element can be found that is not) and also the mentioned condition that makes a relation a function is satisfied vacuously. Conclusion: $\varnothing$ is a function $\varnothing\to B$. No matter whether $B$ is empty or not. An empty function has an empty range/image, but that is not the same thing as co-domain. $\endgroup$ – Vera Sep 6 '18 at 15:29
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The statement "the empty space is the initial object in the category of topological spaces" has nothing to do with homotopies. It simply means there is a unique continuous map $\emptyset\mapsto Y$ for any space $Y$, namely the empty map. It is continuous because the preimage of every open set is open.

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  • $\begingroup$ But a continuous function is a homotopy, no? $\endgroup$ – Yan King Yin Sep 6 '18 at 6:35
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    $\begingroup$ I'm not sure what you mean. How is an arbitrary continuous function a homotopy? $\endgroup$ – Michal Adamaszek Sep 6 '18 at 6:38
  • $\begingroup$ as per @drhab 's comment above, there exists a homotopy $\varnothing \rightarrow Y$ which is the empty function. $\endgroup$ – Yan King Yin Sep 6 '18 at 13:10
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    $\begingroup$ @YanKingYin You can also define a homotopy $X\to Y$ to be a family of continuous functions $f_t:X\to Y$ where $t$ ranges over $[0,1]$. This in such a way that the function $F:X\times[0,1]\to Y$ prescribed by $(x,t)\mapsto f_t(x)$ is continuous. Then if $X=\varnothing$ for every $t\in[0,1]$ the function $f_t$ is the empty function with domain $\varnothing$ and codomain $Y$. It is a co-called constant homotopy (i.e. $f_t:\varnothing\to Y$ does not depend on $t$). See the comment of Vera on your question where it concerns the fact that $Y$ as codomain is not necessarily empty. $\endgroup$ – drhab Sep 6 '18 at 18:28

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