Prove that $(Y,\mathcal T_1)$ also has the fixed point property

Question

Let $(X,\mathcal T)$ have the fixed point property and let $(Y,\mathcal T_1)$ be a space homeomorphic to $(X, \mathcal T)$. Prove that $(Y,\mathcal T_1)$ also has the fixed point property.

I know that since $(X, \mathcal T)$ and $(Y,\mathcal T_1)$ are homeomorphic, there exists a bijection $f: (X, \mathcal T) \to (Y,\mathcal T_2)$.

Also, if $g:(X, \mathcal T) \to (X, \mathcal T)$ is continuous, there exists an $a \in X$ such that $g(a)=a$.

How do I show that $(Y,\mathcal T_1)$ also has the fixed point property?

I was thinking that maybe we can form some type of composition of $f$ and $g$? But I'm not sure if that will give us anything.

Answer 1

Let $f : Y \to X$ be a homeomorphism, and $h : X \to Y$ its inverse.
Then for any continuous $g : Y \to Y$, we have that \begin{align} f \circ g \circ h : X \to X, x \mapsto f(g(h(x))) \end{align} is continuous as it is a composition of continuous functions. So there exists an $a \in X$ such that $f(g(h(a))) = a$. Then $y = h(a) \in Y$ and \begin{align} y = h(a) = h(f(g(h(a)))) = g(h(a)) = g(y), \end{align} so $g$ has a fixed point.

Answer 2

Hint: The relevant composition is $f^{-1}\circ g\circ f$.

Answer 3

let $g:(Y, \tau_1 )\to (Y,\tau_1)$ be any contnious function and $h : (X, \tau )\to (Y, \tau_1)$ is homeomorphism then define $ f: ( X , \tau) \rightarrow (X, \tau) $ where$ f=h^{-1}\circ g \circ h$ then $f$ has a fixed point

Let $a \in X$ is fixed point of $f$ i,e $f(a)=a$ then

$f(a) =h^{-1} \circ g \circ h(a)=a$

$h^{-1} \circ g \circ h(a)=a$

$g(h(a)=h(a)$

let $ h(a)=b$ then $g(b)=b$

Therefore $b$ is a fixed point of $g$

hence $(Y, \tau_1)$ has the fixed point property