**Puzzle 279**

Here's my attempt:

![tensoring](https://raw.githubusercontent.com/ikshv/categories/master/tensors.png?sanitize=true)

To show that \\(h = f' \circ f\\) and \\(e = g' \circ g\\) (Exercise 278), we need the triangles in the picture (bottom) commute. Here I stuck - while \\(f' \circ f\\) and \\(g' \circ g \\) are natural candidates for \\(h\\) and \\(e\\) respectively, they are not the only ones.

One possible explanation is that we need the morphisms \\(f \otimes g \\) and \\(f' \otimes g'\\) of the \\(C \times C\\) category to be composable, such that \\((f' \otimes g') \circ (f \otimes g) = (h \otimes e)\\), and this in turn implies that \\(h = f'\circ f\\) and \\(e = g'\circ g\\), but I'm waiting for a more formal explanation.

Here's my attempt:

![tensoring](https://raw.githubusercontent.com/ikshv/categories/master/tensors.png?sanitize=true)

To show that \\(h = f' \circ f\\) and \\(e = g' \circ g\\) (Exercise 278), we need the triangles in the picture (bottom) commute. Here I stuck - while \\(f' \circ f\\) and \\(g' \circ g \\) are natural candidates for \\(h\\) and \\(e\\) respectively, they are not the only ones.

One possible explanation is that we need the morphisms \\(f \otimes g \\) and \\(f' \otimes g'\\) of the \\(C \times C\\) category to be composable, such that \\((f' \otimes g') \circ (f \otimes g) = (h \otimes e)\\), and this in turn implies that \\(h = f'\circ f\\) and \\(e = g'\circ g\\), but I'm waiting for a more formal explanation.