Let A and B be finite sets for which |A|=|B| and suppose f: A —> B. Prove that f is injective if and only if f is surjective

$n=\vert A\vert=\vert B\vert\\(1)\;If\;f\;-surjective.\\Let's\;prove\;that\;it's\;also\;injective.\\That\;means\;that\;\\\forall y\in B\;\exists x\in A:\;f(x)=y.\\In\;another\;words:\\for\;every\;y\in B\;exists\;x\in A\\such\;as\;f(x)=y.\\Let\;consider\;y_1,y_2\in B(y_1\neq y_2)\\and\;x_1,x_2\in A\;such\;as\\f(x_1)=y_1\;\;\neq\;\;f(x_2)=y_2\\So,\;x_1\;\neq\;x_2.\\So,\;for\;every\;distinct\;y\in B\;there\;is\;\\unique\;x\in A.\\There\;is\;n\;distinct\;y\;\Rightarrow\;n\;distinct\;x.\\So,\;for\;every\;x\;there\;is\;distinct\;y.\\So,\;function\;is\;injective.\\(2)\;If\;f\;-\;injective.\\Let's\;prove\;that\;it's\;also\;surjective.\\For\;different\;x\in A,\;there\;are\;\\diferrent\;y\in B\;such\;as\;f(x)=y.\\Since\;we\;have\;n\;different\;x\in A\\we\;should\;have\;n\;different\;y\in B.\\Since\;n=\vert B\vert\;every\;y\in B\;has\;x\in A\;\\such\;as\;f(x)=y.\\So,\;function\;is\;surjective.\\Checking\;both\;cases\;when\;one\;condition\\is\;true\;we\;get\;second\;condition\;also\;being\;true.$