[Math] Show that two spaces are not homeomorphic

general-topology

Let $H=[-1,1]\times \{0\}$ and $V=\{0\}\times [-1,0)$ in the plane. Let $T=H \cup V$. Show that $T$ is not homeomorphic to the unit interval $I=[0,1]$.

My idea for this problem is that , if we remove a point from the unit interval , we will be left with at most two connected components, but if we remove the origin from $T$ we will be left with $3$ connected components. Is this enough to prove that $I$ and $T$ are not homeomorphic ? How should I write my answer rigirously?

Any help is appreciated, Thanks !

Best Answer

Yes, your proof is correct, but from your comment above, it seems that the reason why it is correct is not completely clear to you. What you are implicitly using is the following proposition.

Proposition: Let $X$ and $Y$ be topological spaces. Suppose there is $x \in X$ such that $X\setminus\{x\}$ has $m$ connected components. If there is no $y \in Y$ such that $Y\setminus\{y\}$ has $m$ connected components, then $X$ and $Y$ are not homeomorphic.

Note: The conditions we need to check are about the spaces $X\setminus\{x\}$ and $Y\setminus\{y\}$, but our conclusion pertains to $X$ and $Y$.

I have included the proof below, you just need to move your mouse over the second grey box to see it, but I recommend you try to prove it yourself first. If you want a hint, move your mouse over the first grey box.

Hint: If $f : X \to Y$ is a homeomorphism, then for any $A \subseteq X$, $f|_A : A \to f(A)$ is a homeomorphism.

Proof: By way of contradiction, suppose that there is a homeomorphism $f : X \to Y$. Then $f|_{X\setminus\{x\}} : X\setminus\{x\} \to Y\setminus\{f(x)\}$ is a homeomorphism. As the number of connected components is a topological invariant, and $X\setminus\{x\}$ has $m$ connected components, $Y\setminus\{f(x)\}$ has $m$ connected components. This is a contradiction as, by assumption, there is no $y \in Y$ such that $Y\setminus\{y\}$ has $m$ connected components.

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