Composition of cartesian product of functions

Question

When is it true that $ (\phi_1 , \phi_2) \circ (\psi_1 ,\psi_2) = (\phi_1 \circ \psi_1 , \phi_2 \circ \psi_2) $ where $\phi_i, \psi_i, I=1,2$ are functions between arbitrary spaces?

Answer 1

Perhaps I just don't understand your notations, but to me your question doesn't quite make sense. The problem is that it is not clear to me what functions you denote by $(\phi_1,\phi_2)$ and $(\psi_1,\psi_2)$, hence it is not clear that they can be composed. There are two ways that I can make sense of your question, but first let me explain my notations a little bit.

Given two sets $C_1,C_2$, we can form their cartesian product $C_1\times C_2$, which has canonical projections $\pi_i:C_1\times C_2\to C_i$. Now for any two functions $g_i:B\to C_i$ (with the same domain $B$), there is a unique way to define a function $(g_1,g_2):B\to C_1\times C_2$ such that $\pi_i\circ (g_1,g_2)=g_i$, namely, we have to define $(g_1,g_2)(b)=(g_1(b),g_2(b))$.

On the other hand, if $D_1,D_2$ are two sets, and if we denote $\pi'_i:D_1\times D_2\to D_i$ the two canonical projections, then for any two functions $h_i:C_i\to D_i$, there is now a unique way to define a function $h_1\times h_2:C_1\times C_2\to D_1\times D_2$ in such a way that $\pi'_i\circ (h_1\times h_2)=h_i\pi_i$, namely, we have to define $(h_1\times h_2)(c_1,c_2)=(h_1(c_1),h_2(c_2))$.

Now suppose we have also a function $f:A\to B$ and two functions $j_i:D_i\to E_i$. Then using either the universal properties or the explicit formulas I gave above, you can prove the following identities : $$(g_1,g_2)\circ f=(g_1\circ f,g_2\circ f)$$ $$(h_1\times h_2)\circ (g_1,g_2)=(h_1\circ g_1,h_2\circ g_2)$$ $$(j_1\times j_2)\circ (h_1\times h_2)=(j_1\circ h_1)\times (j_2\circ h_2).$$ (If you know about category theory : note that this can be proven using only the universal properties, and thus it holds in any category as soon as all the products involved exists.)

Answer 2

I'm assuming you meant to write $(\phi_1 , \phi_2) \circ (\psi_1 ,\psi_2) = (\phi_1 \circ \psi_1 , \phi_2 \circ \psi_2)$.

That's the definition of composition of cartesian product of functions. Therefore, that's true whenever you can do the compositions $\phi_1\circ\psi_1$ and $\phi_2\circ\psi_2$.

Answer 3

Notations like $(f,g)$ where $f$ and $g$ are functions are kind of tricky.

If $f:X\to Y$ and $g:X\to Z$ then $(f,g)$ can stand for the function $X\to Y\times Z$ that is characterized by $p_Y\circ(f,g)=f$ and $p_Z\circ(f,g)=g$ where $p_Y:Y\times Z\to Y$ and $p_Z:Y\times Z\to Z$ denote the projections.

If $f:Y\to X$ and $g:Z\to X$ then $(f,g)$ can stand for the function $Y\sqcup Z\to X$ that is characterized by $(f,g)\circ i_Y=f$ and $(f,g)\circ i_Z=g$ where $i_Y:Y\to Y\sqcup Z$ and $i_Z:Z\to Y\sqcup Z$ denote the injections.

If $f:X_1\to Y_1$ and $g:X_2\to Y_2$ then $(f,g)$ can stand for the function $X_1\times X_2\to Y_1\times Y_2$ characterized by $p_{Y_1}\circ(f,g)=f\circ p_{X_1}$ and $p_{Y_2}\circ(f,g)=g\circ p_{X_2}$. In this case it is IMHO better to write $f\times g$ instead of $(f,g)$.

If further, and according to that notation $f'\times g':Y_1\times Y_2\to Z_1\times Z_2$ then indeed we have: $$(f'\times g')\circ(f\times f)=(f'\circ f)\times(g'\circ g)$$