Is a product of homotopy is again homotopy? If f and g, f' and g' are homotopic, respectively, then f x f' and g x g' homotopic?
If so, does it lead to the result that product of deformation retract is again deformation retract?
 A: Let's say that $f$ and $g$ are homotopic via $H$, and the same for $f',g',H'$:
$$f,g:X \rightarrow Y$$
$$H: I\times X \rightarrow Y, H(0,x)=f(x), H(1,x)=g(x)$$
$$f',g':X' \rightarrow Y'$$
$$H': I\times X' \rightarrow Y', H'(0,x')=f'(x'), H'(1,x')=g'(x')$$
For a homotopy between $f \times f'$ and $g \times g'$ we need a $T$ such that
$$f \times f', g \times g' : X \times X' \rightarrow Y \times Y'$$
$$T : I \times X \times X' \rightarrow Y \times Y'$$
$$T(0,x,x')=(f\times f')(x,x')=(f(x), f'(x'))$$
$$T(1,x,x')=(g\times g')(x,x')=(g(x), g'(x'))$$
We see that $H \times H' : I \times X \times I \times X' \rightarrow Y \times Y'$ already almost provides us with such a $T$, we just have to map $I$ to $I \times I$. There are many ways to do it, we may take a canonical one: $t \mapsto (t,t)$.
So, define $T(t,x,x')=(H\times H')(t,x,t,x')=(H(t,x), H'(t,x'))$.
It is easy to check that
$$T(0,x,x')=(H(0,x),H'(0,x'))=(f(x),f'(x'))$$
$$T(1,x,x')=(H(1,x),H'(1,x'))=(g(x),g'(x'))$$
Thus, $T$ indeed provides a homotopy between $f \times g$ and $f' \times g'$.

If $A$ is a deformation retract of $X$, and $B$ is a deformation retract of $Y$, that is:


*

*$f: X \rightarrow A$ is a retract and $i_A \circ f$  is homotopic to $id_X$, where $i_A$ is inclusion

*$g: Y \rightarrow B$ is a retract and $i_B \circ g$  is homotopic to $id_Y$, where $i_B$ is inclusion


Then $f\times g : X \times Y \rightarrow A \times B$ is a retract, and by the first part of the answer 
$$(i_A \circ f)\times(i_B \circ g)=(i_A \times i_B)\circ(f \times g) = i_{A \times B} \circ (f \times g)$$
is homotopic to 
$$id_X \times id_Y = id_{X \times Y}$$
Thus, $f \times g$ is a deformation retraction from $X \times Y$ to $A \times B$.
