Assume in a convolutional layer's forward pass we have a $10\times10\times3$ image and five $3\times3\times3$ kernels, then $(10\times10\times3) *( 3\times3\times3\times5)$ has the output of dimensions $8\times8\times5$. Therefore the gradients fed backwards to this convolutional layer also have the dimensions $8\times8\times5$.

When calculating the derivative of loss w.r.t. kernels, the formula is the convolution $input \times \frac{dL}{dZ}$. But if the gradients have dimensions $8\times8\times5$, how is it possible to convolve it with $10\times10\times3$? The gradients have $5$ channels while the input only has $3$.

Since during the forward pass the kernel window does element-wise multiplication and brings the channels down to $1$, do the gradients propagate back to each of the $3$ channels equally? Should the $8\times8\times5$ gradients be reshaped into $8\times8\times1\times5$ and broadcast into $8\times8\times3\times5$ before convolving with the layer input?