Product topology and continuity

Question

Hi guys I'm new to topology and was asked to prove the following, of which I am having troubles with:

Let $F:X \times I \rightarrow Y$ be a continous function. For each $t \in I$ let $f_{t} = F(x,t)$. Prove that $f_{t}$ is continuous.

My attempt is as follows:

Let us first fix a $t_{0} \in I$ and let $\{V_{i}\}$ be a collection of open subsets in $Y$ that contains $F(x,t_{0})$. Since $\{V_{i}\}$ are open then this means that the intersection of which is also open, namely $\cap_{i}V_{i}$ is.

Now we use continuity of $F$ and see that $F^{-1}(\cap_{i}V_{i})$ must be a open set in the topology of $X \times I$. But this topology is generated by the basis whose sets are the of form $U_{X} \times U_{I}$, where $U_{X}$ and $U_{I}$ are open sets in the topologies of $X$ and $I$ respectively.

Then $F^{-1}(\cap_{i}V_{i})$ must be of the same form, i.e. $F^{-1}(\cap_{i}V_{i}) = U_{X_1} \times U_{I_1}$ where $U_{X_1}$ and $U_{T_1}$ are open sets in $X$ and $I$ respectively. But notice that $U_{I_1} = {t_0}$. Then $U_{X_1} \times U_{I_1} = U_{X_1} \times {t_0}$.

Hence for a fixed $t_0 \in I$, $F^{-1}(\cap_{i}V_{i}) = U_{X} \times {t_0}$ and in the case of f, $f_{t_0}^{-1}(\cap_{i}V_{i}) = U_{X}$. Therefore the preimage of open sets under $f$ is open.

I'm sorry guys reading over this I can definitely see some errors. Can anyone provide any tips? I'm very confused on this question. Any feedback will be greatly appreciated!

Answer 1

A function $g:A\to B$ is continuous if for every open subset $V\subset B$ we have $f^{-1}(V)$ is open.

Fix $t_0\in I$ and consider $f_{t_0}:X\to Y$ defined by $f_{t_0}(x)=F(x,t_0)$. Now take some open subset $V\subset Y$. Since $F$ is continuous on $X\times I$, we have that $F^{-1}(V)$ is an open subset of $X\times I$, and so $F^{-1}(V)\cap(X\times\{t_0\})$ is an open subset in the relative topology of $X\times\{t_0\}$. We'll show that $f_{t_0}^{-1}(V)=\pi_1\big(F^{-1}(V)\cap(X\times\{t_0\})\big)$, where $\pi_1$ is the projection on the first component. Once we have shown this, we'll know that $f_{t_0}^{-1}(V)$ is open, since $\pi_1$ is an open map (it maps open sets to open sets).

$x\in f_{t_0}^{-1}(V)\Leftrightarrow f_{t_0}(x)=F(x,t_0)\in V\Leftrightarrow (x,t_0)\in F^{-1}(V)\Leftrightarrow (x,t_0)\in F^{-1}(V)\cap(X\times\{t_0\})\Leftrightarrow x\in\pi_1\big(F^{-1}(V)\cap(X\times\{t_0\})\big)$

Therefore $f_{t_0}^{-1}(V)=\pi_1\big(F^{-1}(V)\cap(X\times\{t_0\})\big)$, so $f_{t_0}^{-1}(V)$ is open, and we conclude $f_{t_0}$ is continuous.

Answer 2

Recall that product topology has the following universal property:

Let $\{X_{i}\mid i\in I\}$ be a family of topological spaces, indexed by the index set $I$. Let $X=\prod_{i\in I}X_{i}$ be the product, equipped with the product topology. Let $\pi_{i}:X\rightarrow X_{i}$ be the canonical projection onto the $i$-th coordinate, i.e., $\pi_{i}(x)=x(i)$, $x\in X$. Let $Y$ be a topological space and $f:Y\rightarrow X$. Then the following are equivalent:

(a) $f$ is continuous.

(b) For each $i\in I$, $\pi_{i}\circ f$ is continuous.

Proof: Clearly $(a)\Rightarrow(b)$ because canonical projection $\pi_i$ is continuous by the very definition of product topology. Suppose that $(b)$ holds. Let $i\in I$ and $U_{i}$ be an open set in $X_{i}$, then $f^{-1}(\pi_{i}^{-1}(U_{i}))=(\pi_{i}\circ f)^{-1}(U_{i})$ which is open in $X$. This shows that $f^{-1}(U)$ is open in $X$ for each $U\in\mathcal{C}$, where $\mathcal{C}=\{\pi_{i}^{-1}(U_{i})\mid i\in I\mbox{ and }U_{i}\mbox{ is open in }X_{i}\}$. Recall that $\mathcal{C}$ is a subbase for the product topology, it follows that $f$ is continuous.

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Now, we go back to your problem. We go to prove the following:

Let $X$ and $I$ be topological spaces. Let $t\in I$ be fixed. Define $\theta:X\rightarrow X\times I$ by $\theta(x)=(x,t)$, then $\theta$ is continuous.

Proof: Let $\pi_{X}$ and $\pi_{I}$ be the canonical projections from $X\times I$ onto $X$ and $I$ respectively. Note that $\pi_{X}\circ\theta=id_{X}$, where $id_{X}:X\rightarrow X$, $id_{X}(x)=x$, which is obviously continuous. Also, $\pi_{Y}\circ\theta=c_{t}$, where $c_{t}:X\rightarrow I$ is the constant map $c_{t}(x)=t$, which is also continuous. By the universal property, it follows that $\theta$ is continuous.

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Finally, notice that your $f_t$ is simply the composition $f_t=F\circ\theta$ of continuous maps, so it is also continuous.