Finite Product of path-connected spaces Problem Finite product of path connected spaces is path connected.
Attempt Suppose $X_1$ and $X_2$ be path connected spaces.
Now, I am using Maps into Products.
Let $X_1×X_2$ isn't path connected. Define a map $f:[a,b]\to X_1×X_2$ by $f(x)=(f_1(x),f_2(x))$.
Where $f_1:[a,b]\to X_1$ and $f_2:[a,b]\to X_2$ are continuous because of path-connected.
Then by maps into products $f$ is continuous.
Since image of path connected space is path connected. Therefore $\operatorname{Im}f$ is path connected.But $X_1×X_2$ isn't path connected. Contradiction .Therefore $X_1×X_2$ is path connected. Since $(X_1×X_2×...×X_{n-1})×X_n$ is homeomorphic with $X_1×X_2×...×X_{n-1}×X_n$. Hence finite product of path connected spaces is path connected.
I know this is true for infinite products.
Is it correct? Any help or suggestions to improve this proof. I don't need other proof.
Thanks!
 A: 
Since image of path connected space is path connected. Therefore $\operatorname{Im}f$ is path connected.But $X_1×X_2$ isn't path connected. Contradiction .

Why do you think that $\operatorname{Im}f=X_1\times X_2$? Because that's the claim here, right? No, this is not true in general, even when both $f_1,f_2$ are surjective. By the way: what is $f_1$ and $f_2$ anyway? Being path connected does not mean being an image of $[0,1]$. A space $X$ is path connected if any two points $x,y\in X$ can be connected via a continuous path $f:[0,1]\to X$, i.e. $f(0)=x$ and $f(1)=y$. Clearly $f$ need not be surjective, and in fact often we won't find such surjective path (e.g. when $X=\mathbb{R}$ or any other noncompact space).

So the real solution is as follows: let $(x,y)$ and $(x',y')$ be two points in $X_1\times X_2$. We want to find a continuous function $f:[0,1]\to X_1\times X_2$ such that $f(0)=(x,y)$ and $f(1)=(x',y')$.
Since $X_1$ and $X_2$ are path connected then there are continuous functions
$$f_1:[0,1]\to X_1$$
$$f_2:[0,1]\to X_2$$
such that
$$f_1(0)=x,\ f_1(1)=x'$$
$$f_2(0)=y,\ f_2(1)=y'$$
Now we define
$$f:[0,1]\to X_1\times X_2$$
$$f(t)=\big(f_1(t), f_2(t)\big)$$
Note that $f$ is the path we are looking for. Both ends are as required. And it is continuous by the universal property of products: a function into product is continuous if and only if all its coordinates are continuous.
In fact, this construction can be generalized to any, even infinite, product. Without referring to associativity or induction. Simply replace $X_1,X_2$ with any collection of spaces $\{X_i\}_{i\in I}$.
