Does functor preserve commuting square?

Question

I have a commuting square in category $D$, and a functor $G: D \to E$. The image of the functor in $E$ is also a square.

My question is "Is the squares in the category $E$ also commuting"?

For example:

If we have $F'f \circ \alpha_a = \alpha_b \circ Ff$ in category $D$, does $GF'f \circ G \alpha_a = G \alpha_b \circ GFf$ in category $E$?

PS: This question maybe stupid. I hope it does, because functors should preserve morphisms' composition. But I am a newcomer to category theory, so I hope to confirm it.

Very thanks.

Answer 1

Yes, and precisely for the reason you mentioned. You could write it out in detail to convince yourself: $$ G(F'f) \circ G(\alpha_a) = G(F'f \circ \alpha_a) = G(\alpha_b \circ Ff) = G(\alpha_b) \circ G(Ff). $$ The first and last equality follow from the functoriality of $G$, and the middle equality is just the commutativity of the original square.