I could find a bunch of Q&As on this website about the following proof:

For the finite product space, $X=X_1 \times X_2 \times X_3 \times ... X_n$,

If each $X_i$ is path connected, then X is also path connected.

I am trying to prove the reverse:

For the finite product space, $X=X_1 \times X_2 \times X_3 \times ... X_n$,

If X is path connected, then each of $X_i$ is path connected.

I could not find this answers in this website. I searched many textbooks as follows:

   James Munkres Topology, 2nd.

   Gamelin T.W., Greene R.E. Introduction to topology

   A. Lahiri A First Course in Algebraic Topology

   John McCleary A First Course in Topology

   Seymour Lipschutz Schaum’s Outline of General Topology Schaum’s Outlines  

   Stephen Willard General Topology

Only one of them, Gamelin T.W., Greene R.E. Introduction to topology (page95, Ex 4(e)), has this proof.

However, the solution says I need to refer to the Ex 9.3, page 90, but I cannot understand the solution of it, page 216.

It says as follows:

Prove that if X is path connected and $f:X \to Y$ is $\color{red}{a\,map}$, then f(X) is path connected.

SOLUTION: If $p=f(x)$ and $q=f(y)$, and $\gamma$ is a path in X from p to q, then $f \circ \gamma $ is a path in f(X) from x to y.

First, I do not know the meaning of "$\color{red}{a\,map}$". Does this mean a continuous function?

Also, I guess something is wrong, since x and y are in X, and p and q are in f(X). The statement in the book is the reverse.

Finally, I appreciate if there is another version of easier proof.


2 Answers 2


Suppose that $X$ is path connected. Let $x_i,y_i\in X_i$. Consider $x_j,y_j\in X_j, i\neq j$, you have a continuous path $c(t)$ between $(x_1,...,x_n)$ and $(y_1,...,y_n)$, the composition $p_i\circ c$ where $p_i:X\rightarrow X_i$ is the canonical projection is a continuous path between $x_i$ and $y_i$.

  • $\begingroup$ Thanks for your clear explanation. I wish I could choose both of two answers, but it is impossible. I am sorry that I could not choose yours. $\endgroup$
    – acubens555
    Jul 24, 2017 at 16:22

Yes a map is often used for "continuous function".

He starts with $p,q \in f[X]$ so he writes them right away as $p= f(x)$ for some $x \in X$ and $q = f(y)$ for some $y \in X$. There is a path from $x$ to $y$ in $X$, so a continuous $\gamma: [0,1] \to X$ with $\gamma(0) = x$ and $\gamma(1)=y$.

Then $f \circ \gamma: [0,1] \to Y$ is continuous as a composition of continuous maps and $(f\circ \gamma)(0) = f(\gamma(0)) = f(x) = p$ and $(f \circ \gamma)(1) = f(\gamma(1))= f(y) = q$ so we have a path from $p$ to $q$, for any two points $p,q \in f[X]$, i.e. $f[X]$ is path-connected.

It cannot be made easier than this, IMHO.

Then apply this fact to the continuous projections for the original statement.

  • $\begingroup$ Thanks for your clear explanation. I can perfectly understand it!! $\endgroup$
    – acubens555
    Jul 24, 2017 at 16:23
  • $\begingroup$ @acubens555 It's just restating what you quoted, really. but glad to help.. $\endgroup$ Jul 24, 2017 at 16:31

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