Biproduct in category

Question

Let $(X\oplus X', \pi_X, \pi_{X'}, \iota_X,\iota_{X'})$ and $(Y \oplus Y', p_{Y'},p_Y, j_Y,j_{Y'})$ biproducts in a category $\mathcal{C}$. In MacLane's book, he defines using the structure of product, the "direct sum" of two morphisms. Given morphisms $f: X \rightarrow Y$ and $g: X' \rightarrow Y'$, we define $f\oplus g: X \oplus X' \rightarrow Y \oplus Y'$ to be the unique morphism such that $p_Y \circ (f\oplus g) = f \circ \pi_X$ and $p_{Y'} \circ (f\oplus g) = g \circ \pi_{X'}$. In the next sentense he says that using the coproduct structure i can conclude that $(f\oplus g) \circ \iota_X = i_Y \circ f$ and $(f\oplus g) \circ \iota_{X'} = j_{Y'} \circ g$. I'm trying to prove this afirmation but i did't yet. This follows just using simple calculations in both equalities? Some idea to help me?

Answer 1

The morphisms $(f\oplus g) \circ \iota_X$ and $i_Y \circ f$ are both arrows $$ X\to Y\oplus Y'=Y\times Y'. $$ Hence, to check that they are equal, it suffices to prove separately that $$ p_{Y}\circ (f\oplus g) \circ \iota_X = p_{Y}\circ i_Y \circ f $$ and $$ p_{Y'}\circ (f\oplus g) \circ \iota_X = p_{Y'}\circ i_Y \circ f. $$ Using the definition of $f\oplus g$ and the biproduct identities ($p_{X}\circ \iota_{X}= id$, $p_{X'}\circ \iota_{X}=0$ etc.), you should be able to prove the above equalities.

Of course, the same kind of reasoning can be applied to prove the other equality, i.e. $(f\oplus g) \circ \iota_{X'} = j_{Y'} \circ g$.